Some Results on Maxima and Minima of Real Functions of Vector Variables: A New Perspective

Abstract

This article presents a method for proving several theorems that enable the determination of the maxima and minima of certain classes of real-valued functions with vector variables, without relying on the classical theory based on partial derivatives. However, the main objective of this work is not to compare this approach with existing methods but, rather, to extend the study of extrema in real functions of vector variables that are not differentiable, as illustrated in Example 1. Each theorem is accompanied by various examples that demonstrate its applicability. The results are based on Theorems 1 and 2, as well as the selection of an appropriate connection between Theorem 2 and the functions to be optimized. Additionally, definitions related to the hierarchy between variables within a given domain are introduced, providing the theoretical framework necessary for the development of the proposed results.

1. Introduction

The study of maxima and minima of multivariable functions is a fundamental problem in calculus and optimization. Traditionally, these problems have been approached using methods based on partial derivatives, the gradient, the Hessian matrix, and Lagrange multipliers in the case of constraints. However, one of the main limitations of these approaches is the requirement that the functions be differentiable in order to apply them.

In this work, we present an alternative method for determining the maxima and minima of certain classes of real functions of vector variables, without relying on classical theory based on partial derivatives. Rather than comparing this new methodology with existing approaches, our primary objective is to extend the study of extrema in vector-variable functions that are not differentiable, as illustrated in Example 1.

To achieve this, we establish an appropriate link between the optimization problem and the mathematical relationship, i.e.,

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\left(a_1+a_2+\cdots +a_n\right)=\lambda \left(a_1^2+a_2^2+\cdots +a_n^2\right)\hfill \\ max\{x_1+x_2+\cdots +x_n\}=r\sqrt{n}\hfill \\ min\{x_1+x_2+\cdots +x_n\}=-r\sqrt{n}\hfill \end{array}\right.,\end{array}$$

(1)

which allows us to obtain significant results in optimization theory without requiring differentiability.

Additionally, we implicitly consider that, when solving the problem expressed as $max(min)f(x,y,z)$, subject to a constraint, a hierarchy is established among the variables depending on how f is defined and the domain of each variable. For example, if our problem is $maxf(x,y,z)=x{y}^2{z}^3$ with $x+y+z=6$ and $0<x$, $0<y$, and $0<z$, the relationship expressed as $z>y>x$ is obtained, meaning that the variable with the highest hierarchy is z, followed by y and, finally, x. On the other hand, if we consider the problem expressed as $maxf(x,y,z)=xyz$, subject to $x+y+z=a$, with $0<x<a$, $0<y<a$, and $0<z<a$, we observe that $x=y=z$, indicating that the variables have the same hierarchy in the given domain.

The mathematical relationship expressed by (1) has been widely used in various areas of mathematics due to its versatility, as shown in [1,2,3]. For instance, in [1], different ways of representing an even number ending in six as the sum of two odd numbers were explored, establishing conditions to ensure that these numbers are prime. Likewise, using an appropriate linear functional ($F:{\mathbb{R}}^3\to \mathbb{R}$), representations of natural numbers in the form of $10A+9$, where $A\in \mathbb{N}$, were obtained, enabling the determination of positive integer solutions to the quadratic equation expressed as $(10x+9)(10y+9)=P$, where P is a natural number ending in one. Moreover, this relationship has been employed to solve certain differential equations, including both ordinary and partial differential equations.

Furthermore, in [4], Equation (1) was used to find integer solutions to quadratic polynomials in two variables that represent specific natural numbers. In particular, to verify the primality of a number ending in one, the following equations were analyzed: (i) $P=(10x+9)(10y+9)$, (ii) $P=(10x+1)(10y+1)$, and (iii) $P=(10x+7)(10y+3)$. It was demonstrated that if these equations have no integer solutions, then P is a prime number.

Finally, in [5], the authors proved the Dvoretzky–Rogers, Schur, and Orlicz theorems within the framework of functional analysis, utilizing the Hahn–Banach theorem and the fundamental relationship represented by (1). To do so, they established an appropriate link between these theorems and the mentioned fundamental relationship, providing a solid conceptual and technical foundation to justify their results.

In this context, previous research has emphasized the importance of establishing a suitable connection between the relationship expressed as (1) and the problem under study. One of the main challenges was to construct an appropriate link with an existing theoretical framework that would allow for the derivation of significant results in optimization.

This article is organized into two main sections. Section 2 presents the fundamental theorems that support our methodology, while Section 3 illustrates various concrete cases that demonstrate the applicability and scope of the obtained results.

The relations in (26), presented below, establish the fundamental basis for the development of this article.

Theorem 1. 

Let $f:{\mathbb{R}}^n\to \mathbb{R}$; then, ${\displaystyle maxf(x_1,x_2,...,x_n)=\sum _{i=1}^n{x}_i}$, subject to the following condition: ${\displaystyle \sum _{i=1}^n{x}_i^2=r^2}$ is $r\sqrt{n}$.

Proof. 

Since ${\displaystyle \sum _{i=1}^n{x}_i=\langle (x_1,x_2,...,x_n),(1,1,...,1)\rangle =\parallel x\parallel ·\sqrt{n}·cos\theta =r\sqrt{n}}$, where $\theta$ is the angle formed by the $x=(x_1,x_2,...,x_n)$ vectors and the $(1,1,...,1)$ vector. Here, the maximum and minimum are obtained when $\theta =0$ and $\theta =\pi$, respectively. □

Theorem 2. 

Let $a_1,\dots ,a_n$ be any real numbers; then, there exists $\lambda \in \mathbb{R}$ such that

$$\sum _{i=1}^n{a}_i=\lambda \sum _{i=1}^n{a}_i^2$$

Proof. 

Let $f:{\mathbb{R}}^n\to \mathbb{R}$ be defined by $f\left(x_1,...,x_n\right)={\sum }_{i=1}^n{a}_i{x}_i$, where f is a linear and continuous functional. Therefore,

$${\mathbb{R}}^n=kerf\oplus {(kerf)}^{\perp },$$

(2)

as stated in the orthogonal decomposition theorem (see [6]).

Thus we have

$$dim{\mathbb{R}}^n=dimImf+dimkerf$$

$$n=1+dimkerf$$

Therefore, $dimkerf=n-1$.

• Thus, from (2), we have

$$(1,1,\dots ,1)=\sum _{i=1}^{n-1}{\lambda }_i{u}_i+{\lambda }_n{u}_n,$$

(3)

where $\left\{u_1,\dots ,u_{n-1}\right\}\subset kerf$ and $u_n\in {(kerf)}^{\perp }$.

From (3) and taking into account that f is a linear functional, we have

$$\begin{array}{c}f(1,\dots ,1)=\lambda f\left(u_n\right)\\ {\displaystyle \sum _{i=1}^n}{a}_i=\lambda {\displaystyle \sum _{i=1}^n}{a}_i^2\\ \mathrm{since}{u}_n=\left(a_1,\dots ,a_n\right)\in {(kerf)}^{\perp }\end{array}$$

The proof that $\lambda$ is unique is trivial. □

2. Main Results

Using the aforementioned technique, we must standardize the resolution of various problems.

Theorem 3. 

Let $H:[a,b]\to \mathbb{R},T:[c,d]\to \mathbb{R}$, $G:[a,b]\times [c,d]⟶\mathbb{R}andF(x,y)=H\left(x\right)+G(x,y)+T\left(y\right)$ be continuous functions. Then, $MaxF(x,y)={\displaystyle \frac{3}{\lambda }}$ for some $\lambda >0$, where

$$\begin{array}{c}\hfill MaxH\left(x\right)=MaxT\left(y\right)=MaxG(x,y)={\displaystyle \frac{1}{\lambda }}or\\ \hfill MaxF(x,y)=0,whereMaxH\left(x\right)=MaxT\left(y\right)=MaxG(x,y)=0.\end{array}$$

Proof. 

Let

$$A=H\left(x\right),B=G(x,y),C=T\left(y\right).$$

(4)

Then, using the following relation:

$$A+B+C=\lambda \left[A^2+B^2+C^2\right]$$

we obtain the following:

$$H\left(x\right)+G(x,y)+T\left(y\right)=\lambda \left(H^2\left(x\right)+G^2(x,y)+T^2\left(y\right)\right],$$

(5)

where $\lambda =\lambda (x,y)$.

From the relationship expressed by (5), we obtain

$${\left(H\left(x\right)-{\displaystyle \frac{1}{2\lambda }}\right)}^2+{\left(G(x,y)-{\displaystyle \frac{1}{2\lambda }}\right)}^2+{\left(T\left(y\right)-{\displaystyle \frac{1}{2\lambda }}\right)}^2={\displaystyle \frac{3}{4{\lambda }^2}}.$$

(6)

From Theorem (1), together with the relation expressed by (6), we obtain

$$Max\left(H\left(x\right)+G(x,y)+T\left(y\right)-{\displaystyle \frac{3}{2\lambda }}\right)={\displaystyle \frac{3}{2\left|\lambda \right|}}.$$

(7)

From the relationship expressed by (6), we obtain

$$\begin{array}{c}\hfill \left\{\begin{array}{c}H\left(x\right)-{\displaystyle \frac{1}{2\lambda }}={\displaystyle \frac{\sqrt{3}}{2\left|\lambda \right|}}{b}_1\hfill \\ G(x,y)-{\displaystyle \frac{1}{2\lambda }}={\displaystyle \frac{\sqrt{3}}{2\left|\lambda \right|}}{b}_2\hfill \\ T\left(y\right)-{\displaystyle \frac{1}{2\lambda }}={\displaystyle \frac{\sqrt{3}}{2\left|\lambda \right|}}{b}_3\hfill \end{array},\right.\end{array}$$

(8)

where $b_1=b_1(x,y),b_2=b_2(x,y),b_3=b_3(x,y),\mathrm{and}{b}_1^2+b_2^2+b_3^2=1$. The maximum is reached in (7) when $b_1=b_2=b_3=\pm {\displaystyle \frac{1}{\sqrt{3}}}.$ For $b_1=b_2=b_3={\displaystyle \frac{1}{\sqrt{3}}}$ and $\lambda >0$, we obtain the following from (8):

$$H\left(x\right)=G(x,y)=T\left(y\right)={\displaystyle \frac{1}{\lambda }}$$

(9)

For $\lambda <0$, we obtain

$$H\left(x\right)=G(x,y)=T\left(y\right)=0$$

(10)

Similarly, for $b_1=-{\displaystyle \frac{1}{\sqrt{3}}},b_2=b_3={\displaystyle \frac{1}{\sqrt{3}}}$, we obtain the following from (8):

For $\lambda >0:H\left(x\right)=0,G(x,y)={\displaystyle \frac{1}{\lambda }},T\left(y\right)={\displaystyle \frac{1}{\lambda }}$.

For $\lambda <0:H\left(x\right)={\displaystyle \frac{1}{\lambda }},G(x,y)=0,T\left(y\right)=0$.

The other cases are similar, and the only relations that satisfy the relation expressed by (7) are the relations expressed by (9) and (10). □

Theorem 4. 

Let $V=f_1\left(x\right)f_2\left(y\right)f_3\left(z\right)$, where V is constant and $f_i>0$; then, the minimum of the function expressed as

$$S(x,y,z)=f_1\left(x\right)f_2\left(y\right)+2f_1\left(x\right)f_2\left(y\right)+2f_2\left(y\right)f_3\left(z\right)$$

is given by

$$MinS=3\sqrt[3]{4V^2}.$$

Proof. 

Taking into account the expression of V,

$$S=V\left[{\displaystyle \frac{1}{f_3\left(z\right)}}+{\displaystyle \frac{2}{f_2\left(y\right)}}+{\displaystyle \frac{2}{f_1\left(x\right)}}\right].$$

(11)

Linking this equality with the relationship expressed as

$$a+b+c=\lambda \left[a^2+b^2+c^2\right]$$

(12)

where $a={\displaystyle \frac{1}{f_3\left(z\right)}},b={\displaystyle \frac{2}{f_2\left(y\right)}},c={\displaystyle \frac{2}{f_1\left(x\right)}}$, the following is obtained from the relationship expressed by (12):

$${\left({\displaystyle \frac{1}{f_3\left(z\right)}}-{\displaystyle \frac{1}{2\lambda }}\right)}^2+{\left({\displaystyle \frac{2}{f_2\left(y\right)}}-{\displaystyle \frac{1}{2\lambda }}\right)}^2+{\left({\displaystyle \frac{2}{f_1\left(x\right)}}-{\displaystyle \frac{1}{2\lambda }}\right)}^2={\displaystyle \frac{3}{4{\lambda }^2}}$$

(13)

where $\lambda >0,\lambda =\lambda (x,y,z)$.

Parameterizing the relationship expressed by (13), we obtain

$${\displaystyle \frac{1}{f_3\left(z\right)}}={\displaystyle \frac{\sqrt{3}{b}_1+1}{2\lambda }},{\displaystyle \frac{2}{f_2\left(y\right)}}={\displaystyle \frac{\sqrt{3}{b}_2+1}{2\lambda }},{\displaystyle \frac{2}{f_1\left(x\right)}}={\displaystyle \frac{\sqrt{3}{b}_3+1}{2\lambda }},$$

(14)

where $b_1^2+b_2^2+b_3^2=1$ and $b_i=b(x,y,z)$.

According to Theorem (1), applying to the relation expressed as (13), we obtain

$$Max\left({\displaystyle \frac{1}{f_3\left(z\right)}}-{\displaystyle \frac{1}{2\lambda }}+{\displaystyle \frac{2}{f_2\left(y\right)}}-{\displaystyle \frac{1}{2\lambda }}+{\displaystyle \frac{2}{f_1\left(x\right)}}-{\displaystyle \frac{1}{2\lambda }}\right)=Max\left({\displaystyle \frac{1}{f_3\left(z\right)}}+{\displaystyle \frac{2}{f_2\left(y\right)}}+{\displaystyle \frac{2}{f_1\left(x\right)}}-{\displaystyle \frac{3}{2\lambda }}\right)={\displaystyle \frac{3}{2\lambda }}.$$

(15)

It can be deduced from the relation expressed by (14) that the maximum is reached (15) when $b_1=b_2=b_3={\displaystyle \frac{1}{\sqrt{3}}}$. Therefore, in (14), we hasve the following:

$$f_3\left(z\right)=\lambda ,f_2\left(y\right)=2\lambda ,f_1\left(x\right)=2\lambda$$

(16)

Thus, we have $V=4{\lambda }^3$, which implies that $\lambda =\sqrt[3]{{\displaystyle \frac{V}{4}}}$. Using this fact and the relation expressed as (16) in (11), we obtain the following:

$$\mathrm{Min}S=V\left[{\displaystyle \frac{1}{\lambda }}+{\displaystyle \frac{2}{2\lambda }}+{\displaystyle \frac{2}{2\lambda }}\right]={\displaystyle \frac{6V}{2\lambda }}={\displaystyle \frac{3V}{\lambda }}=3\sqrt[3]{4V^2}$$

The minimum point is obtained from the relation expressed by (16) and the value of $\lambda =3\sqrt[3]{4V^2}$. □

Theorem 5. 

Let $g:[a,b]\to \mathbb{R}$ and $F:[a,b]\times [c,d]\to \mathbb{R}$ be continuous functions; then, there exists $\lambda \in \mathbb{R}$ such that the maximum of the function expressed as $H(x,y)=g\left(x\right)F(x,y)$ occurs when

$$\begin{array}{c}\hfill \left|F(x,y)\right|=1andln\left|g\left(x\right)\right|=-{\displaystyle \frac{1}{\lambda }}or\\ \hfill ln\left|F(x,y)\right|={\displaystyle \frac{1}{\lambda }}and\left|g\left(x\right)\right|=1.\end{array}$$

Proof. 

Let $A=ln\left|F\right(x,y\left)\right|,B=-ln\left|g\right(x\left)\right|$; then, using the relation expressed as

$$A+B=\lambda \left[A^2+B^2\right],$$

we obtain

$$ln\left|F(x,y)\right|-ln\left|g\left(x\right)\right|=\lambda \left[{ln}^2\left|F(x,y)\right|+{ln}^2\left|g\left(x\right)\right|\right],$$

(17)

where $\lambda =\lambda (x,y)$. Then, from the relation expressed as (17), we obtain

$${\left(ln\left|F(x,y)\right|-{\displaystyle \frac{1}{2\lambda }}\right)}^2+{\left(ln\left|g\left(\lambda \right)\right|+{\displaystyle \frac{1}{2\lambda }}\right)}^2={\displaystyle \frac{1}{{2\left|\lambda \right|}^2}}.$$

(18)

Using Theorem (1) in (18), we obtain

$$Max\left(ln\left|F(x,y)\right|-{\displaystyle \frac{1}{2\lambda }}+ln\left|g\left(x\right)\right|+{\displaystyle \frac{1}{2\lambda }}\right)=Max(ln|F(x,y)\left|\right|g\left(x\right)\left|\right)={\displaystyle \frac{1}{\left|\lambda \right|}}.$$

(19)

From the relationship expressed as (17) we obtain the following:

$$\begin{array}{cc}\hfill & ln\left|F(x,y)\right|-{\displaystyle \frac{1}{2\lambda }}={\displaystyle \frac{\sqrt{2}}{2\left|\lambda \right|}}{b}_1\hfill \\ \hfill & ln\left|g\left(x\right)\right|+{\displaystyle \frac{1}{2\lambda }}={\displaystyle \frac{\sqrt{2}}{2\left|\lambda \right|}}{b}_2,\hfill \end{array}$$

(20)

where $b_1^2+b_2^2=1,b_1=b_1(x,y),b_2=b_2(x,y)$.

The maximum reached in (19) is obtained when

$$b_1=b_2=\pm {\displaystyle \frac{1}{\sqrt{2}}}.$$

(21)

For $b_1=b_2=+{\displaystyle \frac{1}{\sqrt{2}}}$ and $\lambda >0$, we obtain the following from (20):

$$ln\left|F(x,y)\right|={\displaystyle \frac{1}{\lambda }}\mathrm{and}ln\left|g\left(x\right)\right|=0.$$

(22)

For $b_1=b_2={\displaystyle \frac{1}{\sqrt{2}}}$ and $\lambda <0$, from the relation expressed by (15), we find that

$$ln\left|F(x,y)\right|=0\mathrm{and}ln\left|g\left(x\right)\right|={\displaystyle \frac{-1}{\lambda }}.$$

(23)

The relations expressed by (22) and (23) verify the relation.

For $b_1={\displaystyle \frac{1}{\sqrt{2}}},b_2=-{\displaystyle \frac{1}{\sqrt{2}}}$ and $\lambda <0$, we obtain the following from (20):

$$ln\left|F(x,y)\right|={\displaystyle \frac{1}{\lambda }}\mathrm{and}ln\left|g\left(x\right)\right|=-{\displaystyle \frac{1}{\lambda }}$$

(24)

For $b_1={\displaystyle \frac{1}{\sqrt{2}}},b_2=-{\displaystyle \frac{1}{\sqrt{2}}}$ and $\lambda <0$, we obtain the following from (20):

$$ln\left|F\right(x,y\left)\right|=0\mathrm{and}ln\left|g\right(x\left)\right|=0$$

(25)

Of the relations expressed by (24) and (25), none verifies the relation expressed by (19). Similar analysis holds for the other cases of $b_1,b_2$, and $\lambda$.

The maximum points are found by solving the systems that appear in the thesis of this theorem, that is,

$$\begin{array}{ccc}\hfill \left|F\right(x,y\left)\right|& =& 1\mathrm{and}ln\left|g\left(x\right)\right|=-{\displaystyle \frac{1}{\lambda }}\mathrm{or}\hfill \\ \hfill ln\left|F\right(x,y\left)\right|& =& {\displaystyle \frac{1}{\lambda }}\mathrm{and}\left|g\left(x\right)\right|=1.\hfill \end{array}$$

□

Theorem 6. 

Let $F(x,y,z)=f\left(x\right)g^3\left(y\right)h^2\left(z\right)$ be a continuous function, where $f,g$, and h are real functions of a real variable. If $f\left(x\right)+g\left(y\right)+h\left(z\right)=a$, where $a>0$, $f\left(x\right)>0$, $g\left(y\right)>0$, and $h\left(z\right)>0$, then the maximum of the function expressed as $F(x,y,z)$ is $MaxF(x,y,z)={\displaystyle \frac{a^6}{432}}$ and is reached when $f\left(x\right)={\displaystyle \frac{a}{6}}$, $g\left(y\right)={\displaystyle \frac{a}{2}}$, and $h\left(z\right)={\displaystyle \frac{a}{3}}$.

Proof. 

Let $A=f\left(x\right)$, $B=g\left(y\right)$, and $C=h\left(z\right)$. Using the relationship expressed as $A+B+C=\lambda [A^2+B^2+C^2]$, we obtain

$$\begin{array}{c}\hfill f\left(x\right)+g\left(y\right)+h\left(z\right)=\lambda [f^2\left(x\right)+g^2\left(y\right)+h^2\left(z\right)],\end{array}$$

(26)

where $\lambda >0$, $\lambda =\lambda (x,y,z)$, and $f\left(x\right)<h\left(z\right)<g\left(y\right)$ in order to obtain the desired maximum of F.

From the relationship expressed as (26) and the problem data, we obtain the following:

$$\begin{array}{c}\hfill a=\lambda \left[a^2-2f\left(x\right)g\left(y\right)-2(a-f\left(x\right)-g\left(y\right))(f\left(x\right)+g\left(y\right))\right].\end{array}$$

(27)

Let

$$\begin{array}{c}\hfill U(x,y)={\displaystyle \frac{f\left(x\right)+g\left(y\right)}{2}},V(x,y)={\displaystyle \frac{g\left(y\right)-f\left(x\right)}{2}}.\end{array}$$

(28)

Replacing (28) in (27), we have

$$\begin{array}{c}\hfill {\displaystyle \frac{a}{2\lambda }}-{\displaystyle \frac{a^2}{6}}=V^2(x;y)+3{\left(U(x;y)-{\displaystyle \frac{a}{3}}\right)}^2.\end{array}$$

(29)

From the relation expressed by (26), we obtain

$$\begin{array}{c}\hfill Max\lambda ={\displaystyle \frac{3}{a}}\end{array}$$

(30)

From the relation of $f\left(x\right)<h\left(z\right)<g\left(y\right)$ and the restriction of $f\left(x\right)+g\left(y\right)+h\left(z\right)=a$, we obtain

$$\begin{array}{c}\hfill f\left(x\right)<{\displaystyle \frac{a}{3}}\mathrm{and}g\left(y\right)>{\displaystyle \frac{a}{3}}.\end{array}$$

(31)

On the other hand, from (29), we obtain $Max{V}^2(x,y)$ when

$$\begin{array}{c}\hfill U(x,y)={\displaystyle \frac{a}{3}}\end{array}$$

(32)

Replacing this last equality in (28), we have

$$\begin{array}{c}\hfill f\left(x\right)+g\left(y\right)={\displaystyle \frac{2a}{3}},\end{array}$$

(33)

and from here, together with the relationship expressed as $f\left(x\right)+g\left(y\right)+h\left(z\right)=a$, we find that

$$\begin{array}{c}\hfill h\left(z\right)={\displaystyle \frac{a}{3}}.\end{array}$$

(34)

From the relation expressed as (33), we obtain the following based on the relation expressed as (31):

$$\begin{array}{c}\hfill g\left(y\right)-{\displaystyle \frac{a}{3}}={\displaystyle \frac{a}{3}}-f\left(x\right)\end{array}$$

(35)

Using Theorem 5, we see that the maximum of the function expressed as $f\left(x\right)\left[-1+{\displaystyle \frac{a}{3f\left(x\right)}}\right]$ is given by

$$\begin{array}{c}\hfill \left|f\left(x\right)\right|=1\mathrm{or}\left|-1+{\displaystyle \frac{a}{3f\left(x\right)}}\right|=1.\end{array}$$

(36)

From the relation expressed as (36), we obtain $-1+{\displaystyle \frac{a}{3f\left(x\right)}}=1$, and from the last relation, we obtain the following:

$$\begin{array}{c}\hfill f\left(x\right)={\displaystyle \frac{a}{6}}.\end{array}$$

(37)

From the relations expressed as (35) and (37), we obtain

$$\begin{array}{c}\hfill g\left(y\right)={\displaystyle \frac{a}{2}}.\end{array}$$

(38)

Therefore, from the relations expressed as (34), (37), and (38) we obtain

$$\begin{array}{c}\hfill MaxF(x,y,z)={\displaystyle \frac{a}{6}}\times {\displaystyle \frac{a^3}{8}}\times {\displaystyle \frac{a^2}{9}}={\displaystyle \frac{a^6}{432}}\end{array}$$

Finally, since the $f\left(x\right)$, $g\left(y\right)$, and $h\left(z\right)$ functions are given, the minimum point is obtained by solving $f\left(x\right)={\displaystyle \frac{a}{6}}$, $g\left(y\right)={\displaystyle \frac{a}{2}}$, and $h\left(z\right)={\displaystyle \frac{a}{3}}$, respectively. □

Remark 1. 

In (35), note that $g\left(y\right)-{\displaystyle \frac{a}{3}}={\displaystyle \frac{g\left(y\right)}{3}}\left(3-{\displaystyle \frac{a}{g\left(y\right)}}\right)$; then, we can apply Theorem 5. Thus,

$$Max\left\{{\displaystyle \frac{g\left(y\right)}{3}}\left(3-{\displaystyle \frac{a}{g\left(y\right)}}\right)\right\}$$

occurs when $\left|{\displaystyle \frac{g\left(y\right)}{3}}\right|=1$ or $\left|3-{\displaystyle \frac{a}{g\left(y\right)}}\right|=1$, obtaining $g\left(y\right)={\displaystyle \frac{a}{2}}$. The expression of ${\displaystyle \frac{a}{4}}$ is discarded, since $g\left(y\right)>{\displaystyle \frac{a}{3}}$.

Theorem 7. 

For $\overline{x}\in {\mathbb{R}}^3$, the extrema of function $F\left(\overline{x}\right)=f_1^2\left(x\right)+f_2^2\left(y\right)+f_3^2\left(z\right)$, where $f_1\left(x\right)$, $f_2\left(y\right)$, and $f_3\left(z\right)$ are continuous functions, subject to the condition of

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\displaystyle \frac{f_1^2\left(x\right)}{A^2}}+{\displaystyle \frac{f_2^2\left(y\right)}{B^2}}+{\displaystyle \frac{f_3^2\left(z\right)}{C^2}}=1,0<A<B<C\hfill \\ f_1\left(x\right)+f_2\left(y\right)=f_3\left(z\right),\hfill \end{array}\right.\end{array}$$

are given by

$$\begin{array}{c}\hfill MaxF\left(\overline{x}\right)={\displaystyle \frac{2A^2{B}^2\left(3C^2+1\right)}{4A^2{B}^2+B^2{(C-1)}^2+A^2{(C+1)}^2}},MinF\left(\overline{x}\right)={\displaystyle \frac{26A^2{B}^2{C}^2}{16B^2{C}^2+9A^2{C}^2+A^2{B}^2}}.\end{array}$$

Proof. 

Let

$$\begin{array}{c}\hfill {\tilde{f}}_1\left(x\right)={\displaystyle \frac{f_1\left(x\right)}{A}},{\tilde{f}}_2\left(y\right)={\displaystyle \frac{f_2\left(y\right)}{B}},{\tilde{f}}_3\left(z\right)={\displaystyle \frac{f_3\left(z\right)}{C}}.\end{array}$$

(39)

From the expression of (39) under the conditions of the problem, we have

$$\begin{array}{c}\hfill F\left(\overline{x}\right)=A^2{\tilde{f}}_1^2\left(x\right)+B^2{\tilde{f}}_2^2\left(x\right)+C^2{\tilde{f}}_3^2\left(z\right),\end{array}$$

(40)

and the constraints are written as

$$\begin{array}{c}\hfill {\tilde{f}}_1^2\left(x\right)+{\tilde{f}}_2^2\left(x\right)+{\tilde{f}}_3^2\left(z\right)=1yA{\tilde{f}}_1\left(x\right)+B{\tilde{f}}_2\left(y\right)=C{\tilde{f}}_3\left(z\right).\end{array}$$

(41)

Let us assume that $f_1\left(x\right)$, $f_2\left(y\right)$, and $f_3\left(z\right)$ are positive; then according to the relation expressed as (41), suppose that

$$\begin{array}{c}\hfill A{\tilde{f}}_1\left(x\right)<B{\tilde{f}}_2\left(y\right)\end{array}$$

(42)

From the relations expressed as (41) and (42), we obtain

$$\begin{array}{c}\hfill B{\tilde{f}}_2\left(y\right)-{\displaystyle \frac{c}{2}}{\tilde{f}}_3\left(z\right)={\displaystyle \frac{c}{2}}{\tilde{f}}_3\left(z\right)-A\widetilde{f_1}\left(x\right).\end{array}$$

(43)

Based on the relation expressed as (43) and using Theorem 5, we have

$$\begin{array}{c}\hfill 2A{\tilde{f}}_1\left(x\right)=(C-1){\tilde{f}}_3\left(z\right),C\ne 1.\end{array}$$

(44)

From the relations expressed as (44) and (41), we obtain

$$\begin{array}{c}\hfill {\tilde{f}}_2\left(y\right)={\tilde{f}}_3\left(z\right){\displaystyle \frac{[C+1]}{2B}}\end{array}$$

(45)

From the relations expressed as (44) and (45) in (41), we obtain

$$\begin{array}{c}\hfill {\tilde{f}}_3\left(z\right)={\displaystyle \frac{4A^2{B}^2}{B^2{(C-1)}^2+A^2{(C+1)}^2+A^2{B}^2}}\end{array}$$

(46)

From the relations expressed as (44), (45), and (46) in (40), we obtain

$$\begin{array}{c}\hfill MaxF\left(\overline{x}\right)={\displaystyle \frac{2A^2{B}^2\left(3C^2+1\right)}{4A^2{B}^2+B^2{(C-1)}^2+A^2{(C+1)}^2}}\end{array}$$

(47)

Of all the possible variants on the signs of the ${\tilde{f}}_1\left(x\right)$, ${\tilde{f}}_2\left(y\right)$, and ${\tilde{f}}_3\left(z\right)$ functions the following is deduced: for ${\tilde{f}}_1\left(x\right)>0$, ${\tilde{f}}_2\left(y\right)<0$, and ${\tilde{f}}_3\left(z\right)>0$, we have the relation expressed as (41).

$$\begin{array}{c}\hfill A{\tilde{f}}_1\left(x\right)=C{\tilde{f}}_3\left(z\right)-B{\tilde{f}}_2\left(y\right)\end{array}$$

(48)

From the relation expressed as (48), assuming that $C{\tilde{f}}_3\left(z\right)<-B{\tilde{f}}_2\left(y\right)$ we obtain the following from from (48):

$$\begin{array}{c}\hfill {\displaystyle \frac{A}{2}}{\tilde{f}}_1\left(x\right)>C{\tilde{f}}_3\left(z\right)\end{array}$$

(49)

From the relations expressed as (48) and (49), we have

$$\begin{array}{c}\hfill {\displaystyle \frac{A}{2}}{\tilde{f}}_1\left(x\right)-C{\tilde{f}}_3\left(z\right)=-{\displaystyle \frac{A}{2}}{\tilde{f}}_1\left(x\right)-B{\tilde{f}}_2\left(y\right).\end{array}$$

(50)

On the other hand, the relation expressed as (50) can also be written as follows:

$$\begin{array}{c}\hfill C{\tilde{f}}_3\left(z\right)\left[-1+{\displaystyle \frac{A{\tilde{f}}_1\left(x\right)}{2C}}\right]=-{\displaystyle \frac{A}{2}}{\tilde{f}}_1\left(x\right)-B{\tilde{f}}_2\left(y\right)\end{array}$$

(51)

Applying Theorem 5 to the product of $C{\tilde{f}}_3\left(z\right)\left[-1+{\displaystyle \frac{A{\tilde{f}}_1\left(x\right)}{2C}}\right]$, we obtain

$$\begin{array}{c}\hfill {\tilde{f}}_1\left(x\right)={\displaystyle \frac{4C}{A}}{\tilde{f}}_3\left(z\right).\end{array}$$

(52)

From the relations expressed as (52) and (48), we obtain

$$\begin{array}{c}\hfill {\tilde{f}}_2\left(y\right)=-{\displaystyle \frac{3C}{B}}{\tilde{f}}_3\left(z\right)\end{array}$$

(53)

Replacing the relations expressed as (52) and (53) in (41), we obtain

$$\begin{array}{c}\hfill {\tilde{f}}_3\left(z\right)={\displaystyle \frac{A^2{B}^2}{16B^2{C}^2+9A^2{C}^2+A^2{B}^2}}\end{array}$$

(54)

Finally, using the relations expressed as (52)–(54) in (40) gives us

$$\begin{array}{c}\hfill MinF\left(\overline{x}\right)={\displaystyle \frac{26A^2{B}^2{C}^2}{A^2{B}^2+9A^2{C}^2+16B^2{C}^2}}.\end{array}$$

The other assumption ($C{\overline{f}}_3\left(z\right)\ge -B{\overline{f}}_2\left(y\right)$) does not lead to optimal values of $F\left(\overline{x}\right)$.

The maximum and minimum points can be obtained from the following system once the ${\tilde{f}}_1\left(x\right)$, ${\tilde{f}}_2\left(y\right)$, and ${\tilde{f}}_3\left(z\right)$ functions have been determined in Equations (46), (45), and (44), respectively:

$${\tilde{f}}_1\left(x\right)={\displaystyle \frac{f_1\left(x\right)}{A}},{\tilde{f}}_2\left(y\right)={\displaystyle \frac{f_2\left(y\right)}{B}},{\tilde{f}}_3\left(z\right)={\displaystyle \frac{f_3\left(z\right)}{C}}$$

These expressions depend on terms A, B, and C. □

Various examples are illustrated below, which are applications of these theorems. Other problems not related to these theorems are solved following the ideas described in the presented theorems.

Remark 2. 

To find the extremes of a function that is not continuous, it is still possible to use the technique shown in the various theorems and examples. Below, we present an example that could be a starting point for such a study.

3. Examples

This section contains illustrative examples of Theorems 3–7. The following example is significant, as it highlights the advantage of this method by maximizing a non-differentiable function under a given condition.

Example 1 

([7]). Find the maximum of the function expressed as $f(x,y)=[x]+{[y]}^2+[x]{[y]}^2$, subject to the following condition:

$$\begin{array}{c}\hfill 2x+y=6,1\le x,0<y.\end{array}$$

(55)

• Solution.

It is clear that $1\le x<y$. Suppose that $2x<y$; from here, together the relation expressed as (55), we obtain

$$\begin{array}{c}\hfill y-3=3-2x.\end{array}$$

(56)

We use the Theorem 5 in the relation expressed as (56).

$$\begin{array}{c}\hfill Max(y-3)=Max(3-2x)\end{array}$$

(57)

From the relation expressed as (57), we obtain $y-3=\pm 1$, which implies that $y=4$ or $y=2$. Replacing (56), we have the following points: $P_1=(1,4)$ and $P_2=(2,2)$. We discard $P_2$, since $2·2\nless 2$.

Another way to write $y-3$ in (56) is $y-3=3\left(-1+{\displaystyle \frac{y}{3}}\right)$, so

$$\begin{array}{c}\hfill Max3\left(-1+{\displaystyle \frac{y}{3}}\right)=Max(3-2x).\end{array}$$

(58)

From (58), together with Theorem 5, we obtain $-1+{\displaystyle \frac{y}{3}}=\pm 1$, which implies $y=6$, which is false, since $x=0$ is not possible due to the relation expressed as (55).

Of all the possible variants, we have $y=4$ and $x=1$. The other assumption is $y<2x$, which does not lead to optimal values of y. Therefore, $Maxf(x,y)=33$.

Example 2 

([7]). Find the extrema of the function expressed as $Max\left({\displaystyle \frac{x}{a}}+{\displaystyle \frac{y}{b}}\right)$, subject to the following constraint: $x^2+y^2=1$.

• Solution.

According to Theorem 2, we have

$$x+y=\lambda \left[x^2+y^2\right].$$

(59)

Using the condition of the problem, we have

$$x+y=\lambda .$$

(60)

Completing squares in (59), we have

$$x-{\displaystyle \frac{1}{2\lambda }}={\displaystyle \frac{b_1}{\sqrt{2}\left|\lambda \right|}}$$

(61)

$$y-{\displaystyle \frac{1}{2\lambda }}={\displaystyle \frac{b_2}{\sqrt{2}\left|\lambda \right|}},$$

(62)

where $b_1^2+b_2^2=1$.

From the relations expressed as (61) and (62), for $\lambda >0$, we obtain

$${\displaystyle \frac{x}{a}}+{\displaystyle \frac{y}{b}}={\displaystyle \frac{1}{2\lambda }}\left[{\displaystyle \frac{\sqrt{2}{b}_1+1}{a}}+{\displaystyle \frac{\sqrt{2}{b}_2+1}{b}}\right].$$

(63)

The relation expressed as (63) can be written as

$${\displaystyle \frac{x}{a}}+{\displaystyle \frac{y}{b}}={\displaystyle \frac{1}{2\lambda }}\left(\left(\sqrt{2}{b}_1+1,\sqrt{2}{b}_2+1\right)·\left({\displaystyle \frac{1}{a}},{\displaystyle \frac{1}{b}}\right)\right)$$

(64)

$${\displaystyle \frac{x}{a}}+{\displaystyle \frac{y}{b}}={\displaystyle \frac{1}{2\lambda }}\sqrt{4+2\sqrt{2}\left(b_1+b_2\right)}·\sqrt{{\displaystyle \frac{1}{a^2}}+{\displaystyle \frac{1}{b^2}}}cos\theta$$

(65)

Furthermore, from the relations expressed as (61) and (62), we obtain the following:

$${\lambda }^2-1={\displaystyle \frac{b_1+b_2}{\sqrt{2}}}$$

(66)

From the relations expressed as (65) and (66), we obtain the following:

$$Max\left({\displaystyle \frac{x}{a}}+{\displaystyle \frac{y}{b}}\right)=\sqrt{{\displaystyle \frac{1}{a^2}}+{\displaystyle \frac{1}{b^2}}}$$

(67)

Example 3 

([8]). Find the maxima and minima of $h(x;y)=x^2+y^2+z^2$, subject to the following conditions:

$$\begin{array}{c}\hfill {\displaystyle \frac{x^2}{4}}+{\displaystyle \frac{y^2}{5}}+{\displaystyle \frac{z^2}{25}}=1andx+y=z\end{array}$$

• Solution.

We use Theorem 7 with $f_1\left(x\right)=x$, $f_2\left(y\right)=y$, $f_3\left(z\right)=z$, $A=2$, $B=\sqrt{5}$, and $C=5$, with which we have

$$\begin{array}{ccc}\hfill MaxF\left(\overline{x}\right)& =& 10\hfill \\ \hfill MinF\left(\overline{x}\right)& =& 4.4520547945.\hfill \end{array}$$

According to Lagrange’s method, $max\left(x^2+y^2+z^2\right)=10$ and $min\left(x^2+y^2+z^2\right)={\displaystyle \frac{75}{17}}$.

Example 4 

([8]). A rectangular box without a lid must have a volume of 32 cubic units. What must be the dimensions be so that the total surface area is minimal?

• Solution.

 

To solve this problem, we follow what is described in Theorem 4.

 

If x, y, and z are the edges, we have

(i) 

Box volume $V=xyz=32$

(ii) 

Box surface $S=xy+2xz+2yz$

Of these relations, we have the following:

$$S=V\left[{\displaystyle \frac{1}{z}}+{\displaystyle \frac{2}{y}}+{\displaystyle \frac{2}{x}}\right]$$

(68)

In the relation expressed as (68), we use the following technique:

$$a+b+c=\lambda \left[a^2+b^2+c^2\right]$$

In this case, taking $a={\displaystyle \frac{1}{z}}$, $b={\displaystyle \frac{2}{y}}$, and $c={\displaystyle \frac{2}{x}}$, we have

$${\displaystyle \frac{1}{z}}+{\displaystyle \frac{2}{y}}+{\displaystyle \frac{2}{x}}=\lambda \left[{\displaystyle \frac{1}{z^2}}+{\displaystyle \frac{4}{y^2}}+{\displaystyle \frac{4}{x^2}}\right]$$

.

Completing squares, we have

$${\left({\displaystyle \frac{1}{z}}-{\displaystyle \frac{1}{2\lambda }}\right)}^2+{\left({\displaystyle \frac{2}{y}}-{\displaystyle \frac{1}{2\lambda }}\right)}^2+{\left({\displaystyle \frac{2}{x}}-{\displaystyle \frac{1}{2\lambda }}\right)}^2={\displaystyle \frac{3}{4{\lambda }^2}}.$$

For this last relation, we find that $\lambda >0$ and

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{z}}-{\displaystyle \frac{1}{2\lambda }}={\displaystyle \frac{\sqrt{3}}{2\lambda }}{b}_1\hfill \\ \hfill & {\displaystyle \frac{2}{y}}-{\displaystyle \frac{1}{2\lambda }}={\displaystyle \frac{\sqrt{3}}{2\lambda }}{b}_2\hfill \\ \hfill & {\displaystyle \frac{2}{x}}-{\displaystyle \frac{1}{2\lambda }}={\displaystyle \frac{\sqrt{3}}{2\lambda }}{b}_3,\hfill \end{array}$$

(69)

where

$$b_1^2+b_2^2+b_3^2=1.$$

From the relation expressed as (69), we find that z, y, and x must be minimum; therefore, $b_1$, $b_2$, and $b_3$ must have maximum values simultaneously. This happens when $b_1=b_2=b_3={\displaystyle \frac{1}{\sqrt{3}}}$. Then, in (69) we have

$$z=\lambda ,y=2\lambda ,x=2\lambda$$

Substituting these last relations in (i), we have $4{\lambda }^3=32$, that is, $\lambda =2$.

Then, the minimum surface in (ii) is

$$S=32\left[{\displaystyle \frac{1}{2}}+{\displaystyle \frac{2}{4}}+{\displaystyle \frac{2}{4}}\right]$$

$$S=32\times {\displaystyle \frac{3}{2}}=48.$$

Example 5 

([8]). What is the maximum volume of the rectangular parallelepiped that can be inscribed in the ellipsoid expressed as ${\displaystyle \frac{x^2}{9}}+{\displaystyle \frac{y^2}{16}}+{\displaystyle \frac{z^2}{36}}=1$?

• Solution.

The volume of the parallelepiped is

$$V=8xyz,$$

(70)

where $(x,y,z)$ belong to the ellipsoid.

Let

$$x=3\tilde{x},y=4\tilde{y},z=6\tilde{z}$$

(71)

According to (70) and (71), we obtain the following:

$$V=72\times 8\tilde{x}\tilde{y}\tilde{z},$$

(72)

where ${\tilde{x}}^2+{\tilde{y}}^2+{\tilde{z}}^2=1$.

Now, using the relationship expressed as

$$\tilde{x}+\tilde{y}+\tilde{z}=\lambda \left[{\tilde{x}}^2+{\tilde{y}}^2+{\tilde{z}}^2\right],$$

we obtain

$${\left(\tilde{x}-{\displaystyle \frac{1}{2\lambda }}\right)}^2+{\left(\tilde{y}-{\displaystyle \frac{1}{2\lambda }}\right)}^2+{\left(\tilde{z}-{\displaystyle \frac{1}{2\lambda }}\right)}^2={\displaystyle \frac{3}{4{\lambda }^2}}$$

Therefore,

$$\begin{array}{cc}\hfill & \tilde{x}-{\displaystyle \frac{1}{2\lambda }}={\displaystyle \frac{\sqrt{3}{b}_1}{2\lambda }}\hfill \\ \hfill & \tilde{y}-{\displaystyle \frac{1}{2\lambda }}={\displaystyle \frac{\sqrt{3}}{2\lambda }}{b}_2\hfill \\ \hfill & \tilde{z}-{\displaystyle \frac{1}{2\lambda }}={\displaystyle \frac{\sqrt{3}}{2\lambda }}{b}_3\hfill \end{array}$$

where

$$b_1^2+b_2^2+b_3^2=1,$$

(73)

and

$$\tilde{x}={\displaystyle \frac{\sqrt{3}{b}_1+1}{2\lambda }},\tilde{y}={\displaystyle \frac{\sqrt{3}{b}_2+1}{2\lambda }},z={\displaystyle \frac{\sqrt{3}{b}_3+1}{2\lambda }}.$$

(74)

It is observed that $\tilde{x},\tilde{y}$, and $\tilde{z}$ are maximum real values if $b_1$, $b_2$, and $b_3$ are maximum, which happens when $b_1=b_2=b_3={\displaystyle \frac{1}{\sqrt{3}}}$. Therefore,

$$\tilde{x}={\displaystyle \frac{1}{\lambda }},\tilde{y}={\displaystyle \frac{1}{\lambda }},\tilde{z}={\displaystyle \frac{1}{\lambda }}.$$

(75)

From the relations expressed as (73) and (75), we find that

$${\displaystyle \frac{3}{{\lambda }^2}}=1.$$

Replacing this in (72), we find that the maximum volume is

$$V_{max}=72\times 8.{\displaystyle \frac{1}{{\lambda }^3}}={\displaystyle \frac{72\times 8v}{3\sqrt{3}}}={\displaystyle \frac{24\times 8\sqrt{3}}{3}}=64\sqrt{3}{u}^3$$

Example 6 

([8]). Find the distance from point $P_0(a,b,c)$ to the plane of $P:Ax+By+Cz=D$.

• Solution.

Consider

$$d(P_0,Q)=\sqrt{{(x-a)}^2+{(y-b)}^2+{(z-c)}^2},whereQ=(x,y,z)\in P.$$

(76)

Defining $f:{\mathbb{R}}^3\to \mathbb{R}$ as

$$f\left(x_1,x_2,x_3\right)=(x-a)x_1+(y-b)x_2+(z-c)x_3,$$

(77)

we have f. Then, defining $f:{\mathbb{R}}^3\to \mathbb{R}$ as

$$f\left(x_1,x_2,x_3\right)=(x-a)x_1+(y-b)x_2+(z-c)x_3,$$

(78)

we find that f is continuous and

$${\mathbb{R}}^3=kerf\oplus {[kerf]}^{\perp }.$$

(79)

Since $(1,1,1)\in {\mathbb{R}}^3$, we according to (77) and (79),

$$(1,1,1)={\lambda }_{1}f+{\lambda }_2=u+v,u\in kerf,v\in {[kerf]}^{\perp }$$

$$f(1,1,1)=f\left(u\right)+f\left(v\right)=\lambda \left[{(x-a)}^2+{(y-b)}^2+{(z-c)}^2\right],$$

(80)

that is,

$$(x-a)+(y-b)+(z-c)=\lambda \left[{(x-a)}^2+{(y-b)}^2+{(z-c)}^2\right].$$

(81)

From the relation expressed as (81), we obtain

$$\begin{array}{cc}\hfill & x-a-{\displaystyle \frac{1}{2\lambda }}={\displaystyle \frac{\sqrt{3}}{2\left|\lambda \right|}}{b}_1\hfill \\ \hfill & y-b-{\displaystyle \frac{1}{2\lambda }}={\displaystyle \frac{\sqrt{3}}{2\left|\lambda \right|}}{b}_2\hfill \\ \hfill & z-c-{\displaystyle \frac{1}{2\lambda }}={\displaystyle \frac{\sqrt{3}}{2u1}}{b}_3\hfill \end{array}$$

(82)

where $b_1^2+b_2^2+b_3^2=1$. Then, according to the relation expressed as (64), we obtain

$$\begin{array}{cc}\hfill & Ax-aA-{\displaystyle \frac{A}{2\lambda }}={\displaystyle \frac{\sqrt{3}}{2\left|\lambda \right|}}{b}_{1}A\hfill \\ \hfill & By-bB-{\displaystyle \frac{B}{2\lambda }}={\displaystyle \frac{\sqrt{3}}{2\left|\lambda \right|}}{b}_{2}B\hfill \\ \hfill & Cz-cC-{\displaystyle \frac{C}{2\lambda }}={\displaystyle \frac{\sqrt{3}}{2\left|\lambda \right|}}{b}_{3}C\hfill \end{array}$$

(83)

According to the relation expressed as (83), we have

$$Ax+By+Cz-aA-bB-cC={\displaystyle \frac{\sqrt{3}{b}_{1}A+\sqrt{3}{b}_{2}B+\sqrt{3}{b}_{3}C+A+B+C}{2\lambda }}$$

(84)

This is true if $\lambda >0$. Since $Q\in P$, we have

$$Ax+By+Cz=D.$$

(85)

From (84) and (85), we obtain the following:

$$\lambda ={\displaystyle \frac{A\left(\sqrt{3}{b}_1+1\right)+B\left(\sqrt{3}{b}_2+1\right)+C\left(\sqrt{3}{b}_3+1\right)}{2(D-aA-bB-cC)}}$$

(86)

From the relations given in (82), we have

$${(x-a)}^2+{(y-b)}^2+{(z-c)}^2={\displaystyle \frac{{\left(\sqrt{3}{b}_1+1\right)}^2+{\left(\sqrt{3}{b}_2+1\right)}^2+{\left(\sqrt{3}{b}_3+1\right)}^2}{4{\lambda }^2}}$$

(87)

Therefore, according to (87) and (76), we obtain

$$d\left(P_0,Q\right)={\displaystyle \frac{{\left[{(\sqrt{3}{b}_1+1)}^2+{\left(\sqrt{3}{b}_2+1\right)}^2+{\left(\sqrt{3}{b}_3+1\right)}^2\right]}^{1/2}}{2\left|\lambda \right|}}$$

(88)

From the relations expressed as (88) and (86), we have

$$d\left(P_0,Q\right)={\displaystyle \frac{{\left[{\left(\sqrt{3}{b}_1+1\right)}^2+{\left(\sqrt{3}{b}_2+1\right)}^2+{\left(\sqrt{3}{b}_3+1\right)}^2\right]}^{1/2}|D-aA-bB-cC|}{\left|A\left(\sqrt{3}{b}_3+1\right)+B\left(\sqrt{3}{b}_2+1\right)+C\left(\sqrt{3}{b}_3+1\right)\right|}}.$$

(89)

where

$$cos\theta ={\displaystyle \frac{〈(A,B,C),\left(\sqrt{3}{b}_1+1,\sqrt{3}{b}_2+1,\sqrt{3}{b}_3+1\right)〉}{\sqrt{A^2+B^2+C^2}·\sqrt{{\left(\sqrt{3}{b}_1+1\right)}^2+{\left(\sqrt{3}{b}_2+1\right)}^2+{\left(\sqrt{3}{b}_3+1\right)}^2}}}.$$

(90)

From the relations expressed as (89) and (90), we obtain

$$d\left(P_0,Q\right)={\displaystyle \frac{|D-aA-bB-cC|}{\sqrt{A^2+B^2+C^2}|cos\theta |}}$$

(91)

Example 7 

([8]). Find the maximum of $x{y}^2{z}^3$, if $x+y+z=6,0<x,0<y,0<z.$

• Solution.

Using Theorem 6 with $f\left(x\right)=x$, $g\left(y\right)=y$, $h\left(z\right)=z$, and $a=6$, we obtain

$$\begin{array}{c}\hfill MaxF(x,y,z)={\displaystyle \frac{6^6}{432}}=108.\end{array}$$

Example 8 

([7]). Find the extremes of the function expressed as $f(x,y,z)=xyz$, subject to the conditions of $x^2+y^2+z^2=1$ and $x+y=2z$.

• Solution.

Using the technique of $a+b=\lambda \left[a^2+b^2\right]$, we obtain $a=x,b=y$, and

$$x+y=\lambda \left[x^2+y^2\right].$$

(92)

From the data and from (92), we obtain

$$2z=\lambda \left(1-z^2\right)$$

(93)

From the relation expressed as (93), we obtain

$${\left(z+{\displaystyle \frac{1}{\lambda }}\right)}^2-{\displaystyle \frac{1}{{\lambda }^2}}=1.$$

(94)

From the relation expressed as (94), it is easy to see that

$$Minz=1-{\displaystyle \frac{1}{\lambda }}\mathrm{y}Maxz=-{\displaystyle \frac{1}{\lambda }}-1,\lambda \ne 0.$$

This relation leads to nothing, since $\lambda =\infty$ is an absurdity in (92). Using the relation expressed as

$$x+y+z=\lambda \left[x^2-y^2+z^2\right],$$

(95)

we find from the data that $3z=\lambda ,whichimpliesthatz={\displaystyle \frac{1}{3}}$. Then, we have

$$\begin{array}{cc}\hfill & x^2+y^2=1-{\displaystyle \frac{{\lambda }^2}{9}}\hfill \\ \hfill & x+y={\displaystyle \frac{2\lambda }{3}}.\hfill \end{array}$$

(96)

From the relation expressed as (96), we obtain the following:

$$Max(x+y)=\sqrt{1-{\displaystyle \frac{{\lambda }^2}{9}}}·\sqrt{2}={\displaystyle \frac{2\lambda }{3}}$$

(97)

Solving this relation, we obtain $\lambda =\pm \sqrt{3}$.

• If $\lambda =+\sqrt{3}$, we obtain $z={\displaystyle \frac{1}{\sqrt{3}}},x={\displaystyle \frac{1}{\sqrt{3}}},y={\displaystyle \frac{1}{\sqrt{3}}}$. Therefore $max\left\{xyz\right\}={\displaystyle \frac{1}{3\sqrt{3}}}$.
• If $\lambda =-\sqrt{3}$, we obtain $x=-{\displaystyle \frac{1}{\sqrt{3}}},y=-{\displaystyle \frac{1}{\sqrt{3}}},z=-{\displaystyle \frac{1}{\sqrt{3}}}.$ Therefore, $min\left\{xyz\right\}={\displaystyle \frac{-4}{3\sqrt{3}}}$.

Example 9 

([7]). Find the highest point on the surface expressed as

$$z={\displaystyle \frac{8}{3}}{x}^3+4y^3-x^4-y^4.$$

• Solution.

To solve this problem, we use Theorem 5. It is observed that

$$z=x^3\left({\displaystyle \frac{8}{3}}-x\right)+y^3(4-y).$$

(98)

Since x and y are independent variables, according to (98), we have

$$0<x⩽{\displaystyle \frac{8}{3}}y0\le y\le 4.$$

(99)

Applying the technique, we have

$$ln{y}^3+ln(4-y)=\lambda \left[{\left(ln{y}^3\right)}^2+ln(4-y)\right]$$

(100)

From the relation expressed as (100), we obtain

$${\left(lny-{\displaystyle \frac{1}{2\lambda }}\right)}^2+{\left(ln(4-y)-{\displaystyle \frac{1}{2\lambda }}\right)}^2={\displaystyle \frac{1}{2{\lambda }^2}}.$$

(101)

From the relation expressed as (101), for $\lambda >0$, we obtain

$$ln{y}^3-{\displaystyle \frac{1}{2\lambda }}={\displaystyle \frac{\sqrt{2}{b}_1}{2\lambda }},$$

(102)

so

$$ln(4-y)-{\displaystyle \frac{1}{2\lambda }}={\displaystyle \frac{\sqrt{2}{b}_2}{2\lambda }},where{b}_1^2+b_2^2=1,\lambda =\lambda \left(y\right).$$

(103)

The maximum value of $ln{y}^3+ln(4-y)=ln(4-y)y^3$ is obtained when $b_1=b_2=\pm {\displaystyle \frac{1}{\sqrt{2}}}$. Of the four possibilities, we obtain $b_2=-{\displaystyle \frac{1}{\sqrt{2}}}$ and $b_1={\displaystyle \frac{1}{\sqrt{2}}}$. This is true, since $y^3$ and y must have different hierarchies in $y\in \langle 0,4]$.

Therefore, $ln(4-y)=0\iff 4-y=1\iff y=3$. For the x variable, we have

$$ln{x}^3+ln\left({\displaystyle \frac{8}{3}}-x\right)=\tau \left[{\left(ln{x}^3\right)}^2+{ln}^2\left({\displaystyle \frac{8}{3}}-x\right)\right].$$

(104)

From the relation expressed as (104), for $\tau >0$, we have

$$\begin{array}{cc}\hfill & ln{x}^3-{\displaystyle \frac{1}{2\tau }}={\displaystyle \frac{\sqrt{2}{\tilde{b}}_1}{2\tau }}\hfill \\ \hfill & ln\left({\displaystyle \frac{8}{3}}-x\right)-{\displaystyle \frac{1}{2\tau }}={\displaystyle \frac{\sqrt{2}{\tilde{b}}_2}{2\tau }}.\hfill \end{array}$$

(105)

The maximum value of $ln{x}^3+ln\left({\displaystyle \frac{8}{3}}-x\right)$ is obtained when

$${\tilde{b}}_1={\tilde{b}}_2=\pm {\displaystyle \frac{1}{\sqrt{2}}}.$$

For ${\tilde{b}}_1={\displaystyle \frac{1}{\sqrt{2}}},{\tilde{b}}_2=-{\displaystyle \frac{1}{\sqrt{2}}}$, we obtain ${\displaystyle \frac{8}{3}}-x=1$, which implies that $x={\displaystyle \frac{5}{3}}$. Therefore, in (98), we obtain

$$maxz={\displaystyle \frac{12.5}{27}}\left({\displaystyle \frac{8}{3}}-{\displaystyle \frac{5}{3}}\right)+27(4-3)=27+4.62962=31.629$$

(106)

Applying the superior calculus theory, we obtain the following maximum:

$$Maxz=32.333$$

The error that is is $0.7$.

Remark 3. 

Now, consider a case in which, instead of Equation (104), we use the following expression:

$$ln{\displaystyle \frac{2}{3}}{x}^3+ln\left(4-{\displaystyle \frac{3}{2}}x\right)=\tau \left[{ln}^2{\displaystyle \frac{2}{3}}{x}^3+{ln}^2\left(4-{\displaystyle \frac{3}{2}}x\right)\right].$$

Based on this relationship, similar to what was done in (104), we obtain $x=2$. Therefore, we have

$$max\left\{z\right\}={\displaystyle \frac{16}{3}}+27=32.333.$$

Example 10 

([7]). Determine the absolute maximum and minimum of the function expressed as

$$z=sinx+siny+sin(x+y),$$

where $0⩽x⩽\pi /2,0⩽y⩽\pi /2$.

• Solution.

Following what is described in Theorem 3, we have the following:

Using the relationship expressed as

$$A+B+C=\lambda \left[A^2+B^2+C^2\right],$$

(107)

where $A=senx,B=seny,C=sen(x+y),\lambda =\lambda (x,y)$. After replacing these values in the relation expressed as (107), we have

$$sinx+seny+sen(x+y)=\lambda \left[{sin}^{2}x+{sin}^{2}y+{sen}^2(x+y)\right]$$

(108)

From the relation expressed as (108), we have

$$\begin{array}{cc}\hfill & sinx-{\displaystyle \frac{1}{2\lambda }}={\displaystyle \frac{\sqrt{3}{b}_1}{2\lambda }}\hfill \\ \hfill & siny-{\displaystyle \frac{1}{2\lambda }}={\displaystyle \frac{\sqrt{3}{b}_1}{2\lambda }}\hfill \\ \hfill & sin(x+y)-{\displaystyle \frac{1}{2\lambda }}={\displaystyle \frac{\sqrt{3}{b}_3}{2\lambda }},\hfill \end{array}$$

(109)

where $\lambda >0$ and $b_1^2+b_2^2+b_3^2=1$.

Then,

$$Max\left(Senx+Seny+Sen(x+y)\right)=Max{\displaystyle \frac{Cos\theta +3}{2\lambda }}={\displaystyle \frac{3}{\lambda }},$$

(110)

and this value is reached when $b_1=b_2=b_3={\displaystyle \frac{1}{\sqrt{3}}}$; therefore, based on the relation expressed as (109), we have

$$Senx={\displaystyle \frac{1}{\lambda }},Seny={\displaystyle \frac{1}{\lambda }},Cosy+Cosx=1.$$

(111)

From the relation expressed as (111), we obtain

$${\displaystyle \frac{\sqrt{{\lambda }^2-1}}{\lambda }}+{\displaystyle \frac{\sqrt{{\lambda }^2-1}}{\lambda }}=1.$$

(112)

Solving the relation obtained in (112), we have

$$\lambda =\pm {\displaystyle \frac{2}{\sqrt{3}}}$$

(113)

From the relation expressed as (110) y (113), we find that $maxz={\displaystyle \frac{3\sqrt{3}}{2}}$.

Since $x\in [0,\pi /2],y\in [0,\pi /2]$, for $\lambda \le 0$, we find that $maxz=0$, since $senx⩾0,seny⩾0,sen(x+y)⩾0$.

Note: The maximum given in (110) is correct, since x and y have the same hierarchy in the interval of $[0,\pi /2]$

Example 11. 

[7] Find the maxima and minima of the function expressed as $z=x^3+y^3-3xy,0\le x\le 2,-1\le y\le 2$.

• Solution.

We use Theorem 5 as indicated below. Consider $z=x^3+y\left(y^2-3x\right)$. Then we use the following:

$$ln\left|y\right|+ln\left|y^2-3x\right|=\lambda \left[{ln}^2\left|y\right|+{ln}^2\left|y^2-3x\right|\right]$$

$${\left(ln\left|y\right|-{\displaystyle \frac{1}{2\lambda }}\right)}^2+{\left(ln\left|y^2-3x\right|-{\displaystyle \frac{1}{2\lambda }}\right)}^2={\displaystyle \frac{1}{2{\lambda }^2}}$$

(114)

From the relation expressed as (114), for $\lambda >0$, we obtain

$$\begin{array}{cc}\hfill & ln\left|y\right|-{\displaystyle \frac{1}{2y}}={\displaystyle \frac{\sqrt{2}{b}_1}{2\lambda }}\hfill \\ \hfill & ln\left|y^2-3x\right|-{\displaystyle \frac{1}{2\lambda }}={\displaystyle \frac{\sqrt{2}{b}_2}{2\lambda }},\hfill \end{array}$$

(115)

where $b_1^2+b_2^2=1$. According to the relation expressed as (115), $ln\left|y\right|\left|y^2-3x\right|=lny\left(y^2-3x\right)=ln\left|y\right|+ln\left|y^2-3x\right|$, so we have

$$Maxlny\left(y^2-3x\right)={\displaystyle \frac{1}{\lambda }}.$$

(116)

The maxima or minima are reached at the points $b_1=b_2=\pm {\displaystyle \frac{1}{\sqrt{2}}}$; then, we have $ln\left|y\right|=0$. Therefore, $y=\pm 1$. In the relation expressed as (115), we have

$${ln}^2\left|y^2-3x\right|={\displaystyle \frac{1}{\lambda }}=ln|1-3x|.$$

(117)

Accoridng to this last equality,

$$Max{\displaystyle \frac{1}{\lambda }}=ln5.$$

(118)

Therefore, we have (118), and (116)

$$Maxlny\left(y^2-3x\right)=ln5$$

(119)

In addition, $Minln\left|y\right|\left(y^2-3x\right)\mid =0$ occurs when $y=-1$, $\left|y^2-3x\right|=1$, and $x=0$.

Therefore, $Minz=-1$.

Example 12 

([7]). At what point of the ellipse expressed as ${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$ does the line tangent to this line form the triangle of the minor area?

• Solution.

We know that the tangent line to a circle of $x^2+y^2=1$ at the point expressed by $P=\left(x_0,y_0\right)$ is given by $L_T:siny=-{\displaystyle \frac{x_0}{y_0}}x+B$, where the slope is $-{\displaystyle \frac{x_0}{y_0}}$. As $\left(x_0,y_0\right)\in L_T$, we obtain B, that is to say,

$$L_T:y=-{\displaystyle \frac{x_0}{y_0}}x+{\displaystyle \frac{y_0^2+x_0^2}{y_0}}$$

(120)

Por lo tanto, la recta tangente a la elipse en el punto $Q=\left(x_0,y_0\right)$ puede ser hallado usando la transformación

$$x=a\tilde{x},y=b\tilde{y}$$

(121)

Therefore, using the transformation expressed as (121), we obtain the circumference as follows:

$${\tilde{x}}^2+{\tilde{y}}^2=1$$

From the relations expressed as (120) and (121), we find that the equation of the line tangent to the ellipse is given by

$$\tilde{y}={\displaystyle \frac{-{\tilde{x}}_0}{{\tilde{y}}_0}}\tilde{x}+{\displaystyle \frac{{\tilde{x}}_0^2+{\tilde{y}}_0^2}{y_0}}$$

(122)

As $x_0=a{\tilde{x}}_0,y_0=b{\tilde{y}}_0$, we obtain the following from (122):

$${\displaystyle \frac{y}{b}}={\displaystyle \frac{-x_{0}b}{a{y}_0}}{\displaystyle \frac{x}{a}}+{\displaystyle \frac{\frac{y_0^2}{b^2}+\frac{x_0^2}{a^2}}{\frac{y_0}{b}}}$$

(123)

Then, according to Equation (123), we obtain

$$y=-{\displaystyle \frac{b^2}{a^2}}{\displaystyle \frac{x_0}{y_0}}x+{\displaystyle \frac{b^2}{y_0}}.$$

(124)

Therefore, according to Equation (124), we determine the area limited by the tangent line and the coordinate axes as follows:

$$S\left(x_0,y_0\right)={\displaystyle \frac{a^2{b}^2}{2x_0{y}_0}}$$

(125)

Using the described technique, we find that

$$A+B=\lambda \left[A^2+B^2\right].$$

(126)

Now, consider

$$A={\displaystyle \frac{x_0}{a}},B={\displaystyle \frac{y_0}{b}}.$$

(127)

From the relations expressed as (126) and (127), we have

$${\displaystyle \frac{x_0}{a}}+{\displaystyle \frac{y_0}{b}}=\lambda .$$

(128)

From (128), we obtain the squaring as $1+{\displaystyle \frac{2x_0{y}_0}{2b}}={\lambda }^2$, which implies that

$$x_0{y}_0={\displaystyle \frac{ab}{2}}\left({\lambda }^2-1\right)$$

(129)

From the relations expressed as (126) and (127), we obtain

$$\begin{array}{cc}\hfill & {\displaystyle \frac{x_0}{2}}-{\displaystyle \frac{1}{2\lambda }}={\displaystyle \frac{\sqrt{2}b}{2\lambda }},\lambda >0\hfill \\ \hfill & {\displaystyle \frac{y_0}{b}}-{\displaystyle \frac{1}{2\lambda }}={\displaystyle \frac{\sqrt{2}}{2\lambda }}{b}_2,\hfill \end{array}$$

(130)

where $b_1^2+b_2^2=1$.

From the relations expressed as (130) and (128), we have

$$\lambda -{\displaystyle \frac{1}{\lambda }}={\displaystyle \frac{\sqrt{2}}{2\lambda }}\left(b_2+b_1\right)$$

(131)

From the relation expressed as (131), we obtain

$$Max\left({\lambda }^2-1\right)=1$$

(132)

From the relations expressed as (132) and (129), we obtain

$$Max{x}_0{y}_0={\displaystyle \frac{ab}{2}}$$

Example 13 

([7]). The courses of two rivers (within the limits of a determined region) represent an approximate parabola ($y=x^2$) and a straight line ($x-y-2=0$). It is necessary to unite these rivers by means of a rectilinear channel that has the shortest possible length. At what points will it be necessary to draw them?

• Solution.

Let $P=(x,y)$ be a point on the parabola and $Q=(Z,w)$ be a point on the line. The distance from P to Q is given by

$$d(P,Q)=\sqrt{{(x-z)}^2+{(y-w)}^2}.$$

(133)

Therefore, the function to minimize is the one given by Equation (133), subject to the following condition:

$$y=x^{2}1z-w-2=0$$

(134)

Let $a=z-x$ and $b=y-w$; then, using the relation expressed as

$$a+b=\lambda \left[a^2+b^2\right],$$

(135)

we obtain

$$2+x^2-x=\lambda \left[{(z-x)}^2+{(y-\omega )}^2\right].$$

(136)

Furthermore,

$$min\sqrt{{(z-x)}^2+{(y-\omega )}^2}=min\sqrt{{\displaystyle \frac{\left(z+x^2-x\right)·2}{2\lambda }}}.$$

(137)

From the relation expressed as (134), we have

$${\left(z-x-{\displaystyle \frac{1}{2\lambda }}\right)}^2+\left(y-w-{\displaystyle \frac{1}{2\lambda }}\right)={\displaystyle \frac{1}{2{\lambda }^2}}.$$

(138)

From the relation expressed as (138), we have

$$\begin{array}{cc}\hfill & z-x={\displaystyle \frac{1+\sqrt{2}{b}_1}{2\lambda }}\hfill \\ \hfill & y-w={\displaystyle \frac{1+\sqrt{2}{b}_2}{2\lambda }}\hfill \end{array}$$

(139)

where $b_1^2+b_2^2=1$. From the relations in (139), we obtain

$$2+x^2-x={\displaystyle \frac{2+\sqrt{2}\left(b_1+b_2\right)}{2\lambda }}$$

(140)

According to the relations expressed as (138) and (140), we obtain

$$Min\sqrt{{(y-x)}^2+{(y-w)}^2}=Min{\displaystyle \frac{\left(2+x^2-x\right)\sqrt{2}}{\sqrt{2+\sqrt{2}\left(b_1+b_2\right)}}}$$

(141)

The minimum in (141) is given when $b_1={\displaystyle \frac{1}{\sqrt{2}}},b_2={\displaystyle \frac{1}{\sqrt{2}}}$. Therefore,

$$\begin{array}{cc}\hfill Min\sqrt{{(z-x)}^2+{(y-w)}^2}& =Min{\left({\displaystyle \frac{\left(2+x^2-x\right.}{2}}\right)}^{\sqrt{2}}\hfill \\ \hfill & =Min{\displaystyle \frac{\left({\left(x-\frac{1}{2}\right)}^2+\frac{7}{4}\right)}{2}}\sqrt{2}\hfill \\ \hfill & ={\displaystyle \frac{7}{8}}\sqrt{2},\hfill \end{array}$$

(142)

and this happens when $x={\displaystyle \frac{1}{2}}$; therefore, $y={\displaystyle \frac{1}{4}}$. To calculate $(z,w)$ in (142), we obtain

$$\sqrt{{\left(z-{\displaystyle \frac{1}{2}}\right)}^2+{\left({\displaystyle \frac{1}{4}}-(z-2)\right)}^2}={\displaystyle \frac{7}{8}}\sqrt{2}$$

(143)

Solving this equation, we get z.

4. Conclusions

The obtained theorems and illustrated examples demonstrate that the relationship expressed as

$$\left(a_1+a_2+\cdots +a_n\right)=\lambda \left(a_1^2+a_2^2+\cdots +a_n^2\right),$$

where $a_1,a_2,\dots ,a_n\in \mathbb{R}$, is valid for some $\lambda \in \mathbb{R}$. Therefore it can be linked to the problem under study, applying Theorem 1, to obtain the expected results, namely Theorems 3–7.

The presented theorems and examples serve as a starting point for the development of a general theory that facilitates the determination of constrained maxima and minima in real functions of a vector variable, without requiring differentiability. However, a key challenge lies in establishing an appropriate link between the function under study and the relationships given in Theorems 1 and 2.

When addressing optimization problems in functions with a large number of variables, computational difficulties may arise. Therefore, the use of computational methods is recommended for their resolution.

It is also suggested that future research include comparisons between this method and existing approaches, as this study introduces a new perspective for analyzing extrema of real functions of multiple variables. Specifically, the problem is reformulated in the n-dimensional sphere as

$$\left(a_1+a_2+\cdots +a_n\right)=\lambda \left(a_1^2+a_2^2+\cdots +a_n^2\right),$$

leveraging its properties to obtain the desired results.