Investigation of functions for constrained extrema.

The function $u=f(P)=f(x_1, x_2, ..., x_n)$ has a constrained maximum (minimum) at point $P_0(x_1, ..., x_n)$ if there exists a neighborhood of point $P_0$, for all points $P\neq P_0$ in which satisfy the constraint equations $$\varphi_k(P)=\varphi_k(x_1,..., x_n)=0\quad (k=1, 2, ..., m:,,,,, mf(P)$$ (respectively $f(P_0)

The problem of finding a constrained extremum is reduced to investigating the ordinary extremum of the Lagrange function $$L(x_1, x_2, ..., x_n, \lambda_1, ..., \lambda_m)=f(x_1, ..., x_m)+\sum\limits_{k=1}^m\lambda_k\varphi_k(x_1,..., x_n);$$

$\lambda_k\quad (k=1, 2, ..., m)$ are called Lagrange multipliers.

The necessary conditions for a constrained extremum are expressed by a system of $n+m$ equations$$\frac{\partial L(P)}{\partial x_i}=0\quad (i=1, 2, ..., n)\qquad \,\,\,\,\quad{}\\\varphi_k(P)=0\quad (k=1, 2, ..., m)\qquad \quad(1)$$ from which the unknowns $$x_1^0,\, ..., x_n^0,\,\lambda_1^0,\,...,\,\lambda_m^0,$$ can be found, where $x_1^0,, ..., x_n^0$ are the coordinates of the point where a constrained extremum may occur.

The sufficient conditions for a constrained extremum are related to studying the sign of the second differential of the Lagrange function $d^2 L(x_1^0,, ..., x_n^0,,\lambda_1^0,,...,,\lambda_m^0, dx_1,,...,dx_n)$ for each set of values $x_1^0,, ..., x_n^0,,\lambda_1^0,,...,,\lambda_m^0$ obtained from (1) under the condition that $dx_1,, dx_2,, ..., dx_n$ satisfy the equations$$\sum\limits_{j=1}^n\frac{\partial \varphi_k(x_1^0,..., x_n^0)}{\partial x_j}dx_j=0\quad (k=1, 2, ..., m)\qquad (2)$$ where $dx_1^2+dx_2^2+...+dx^2_n\neq 0.$ Specifically, the function $f(P)$ has a constrained maximum at point $P_0(x_1^0,..., x_n^0)$ if for all possible values of $dx_1, ..., dx_n$ satisfying conditions (2) and not all equal to zero simultaneously, the inequality $d^2 L(x_1^0,, ..., x_n^0,,\lambda_1^0,,...,,\lambda_m^0, dx_1,,...,dx_n)<0$ holds, and it has a constrained minimum if under these conditions $d^2 L(x_1^0,, ..., x_n^0,,\lambda_1^0,,...,,\lambda_m^0, dx_1,,...,dx_n)>0.$

In the case of a function $z=f(x, y)$ with the constraint equation $\varphi(x,, y)=0$, the Lagrange function takes the form $L(x, y, \lambda)=f(x, y)+\lambda\varphi(x,y)$.

System (1) consists of three equations:

$$\frac{\partial L}{\partial x}=0,\,\,\,\frac{\partial L}{\partial y}=0,\,\,\,\varphi(x,y)=0.$$

Let $P_0(x_0, y_0), , \lambda_0$ be any solution of this system, and $$\Delta=-\begin{vmatrix}0&\varphi_x'(P_0)&\varphi_y'(P_0)\\\varphi_x'(P_0)&L_{xx}''(P_0,\lambda_0)&L_{xy}''(P_0,\lambda_0)\\\varphi_y'(P_0)&L_{xy}''(P_0,\lambda_0)&L_{yy}''(P_0,\lambda_0)\end{vmatrix}.$$

If $\Delta<0$, then the function $z=f(x, y)$ has a constrained maximum at point $P_0(x_0, y_0)$, and if $\Delta>0$, then it has a constrained minimum.

Examples.

Find the constrained extrema of functions.

1. $z=x^2+y^2-xy+x+y-4$ subject to $x+y+3=0.$

Solution.

Let's form the Lagrange function:

$$L(x, y, \lambda)=x^2+y^2-xy+x+y-4+\lambda(x+y+3).$$

We have $$\frac{\partial L}{\partial x}=2x-y+1+\lambda,\,\,\,\frac{\partial L}{\partial y}=2y-x+1+\lambda.$$

The system (1) takes the form: $$\left\{\begin{array}{lcl}2x-y+1+\lambda=0\\2y-x+1+\lambda=0\\x+y+3=0\end{array}\right.$$

Let's solve the system using Cramer's rule:

$$\Delta=\begin{vmatrix}2&-1&1\\-1&2&1\\1&1&0\end{vmatrix}=-1-1-2-2=-6;$$ $$\Delta_1=\begin{vmatrix}-1&-1&1\\-1&2&1\\-3&1&0\end{vmatrix}=-1+3+6+1=9;$$ $$\Delta_2=\begin{vmatrix}2&-1&1\\-1&-1&1\\1&-3&0\end{vmatrix}=3-1+1+6=9;$$ $$\Delta_3=\begin{vmatrix}2&-1&-1\\-1&2&-1\\1&1&-3\end{vmatrix}=-12+1+1+2+2+3=-3.$$ From here, we find: $$x=\frac{\Delta_1}{\Delta}=\frac{9}{-6}=-1,5;\quad y=\frac{\Delta_2}{\Delta}=\frac{9}{-6}=-1,5;\quad \lambda=\frac{\Delta_3}{\Delta}=\frac{-3}{-6}=0,5.$$

Next, we find the second derivatives: $$\frac{\partial^2 L}{\partial x^2}=\frac{\partial^2 L}{\partial y^2}=2,\,\,\,\frac{\partial^2 L}{\partial x\partial y}=-1;$$ $$\varphi'_x=1,\qquad \varphi'_y=1$$

$$\Delta=-\begin{vmatrix}0&1&1\\1&2&-1\\1&-1&2\end{vmatrix}=-(-1-1-2-2)=6>0.$$ Therefore, at point $P(-1.5, -1.5)$, the function has a constrained minimum.

$z_{min}=(-1.5)^2+(-1.5)^2-(-1.5)^2-1.5-1.5-4=2.25-3-4=-4.75.$

Answer: At point $P(-1.5, -1.5)$, the function has a constrained minimum; $z_{min}=-4.75.$

2. $z=\frac{1}{x}+\frac{1}{y}$ subject to $x+y=2.$

Solution.

Let's form the Lagrange function: $$L(x, y, \lambda)=\frac{1}{x}+\frac{1}{y}+\lambda(x+y-2).$$

Имеем $$\frac{\partial L}{\partial x}=-\frac{1}{x^2}+\lambda,\,\,\,\frac{\partial L}{\partial y}=-\frac{1}{y^2}+\lambda.$$

The system (1) takes the form: $$\left\{\begin{array}{lcl}-\frac{1}{x^2}+\lambda=0\\-\frac{1}{y^2}+\lambda=0\\x+y-2=0\end{array}\right.$$

Let's solve the system:

$$\left\{\begin{array}{lcl}-\frac{1}{x^2}+\lambda=0\\-\frac{1}{y^2}+\lambda=0\\x+y-2=0\end{array}\right.\Rightarrow\left\{\begin{array}{lcl}\lambda=\frac{1}{x^2}\\\lambda=\frac{1}{y^2}\\x=2-y\end{array}\right.\Rightarrow\left\{\begin{array}{lcl}\lambda=\frac{1}{x^2}\\x^2={y^2}\\x=2-y\end{array}\right.\Rightarrow\left\{\begin{array}{lcl}\lambda=\frac{1}{x^2}\\(2-y)^2={y^2}\\x=2-y\end{array}\right.\Rightarrow$$

$$\Rightarrow\left\{\begin{array}{lcl}\lambda=\frac{1}{x^2}\\4-4y+y^2={y^2}\\x=2-y\end{array}\right.\Rightarrow\left\{\begin{array}{lcl}\lambda=1\\y=1\\x=1\end{array}\right.$$

Next, we find the second derivatives: $$\frac{\partial^2 L}{\partial x^2}=\frac{2}{x^3};\qquad \frac{\partial^2 L}{\partial y^2}=\frac{2}{y^2},\,\,\,\frac{\partial^2 L}{\partial x\partial y}=0;$$ $$\varphi'_x=1,\qquad \varphi'_y=1$$

$$\frac{\partial^2 L(x_0, y_0)}{\partial x^2}=2;\qquad \frac{\partial^2 L(x_0, y_0)}{\partial y^2}=2,\,\,\,\frac{\partial^2 L}{\partial x\partial y}=0;$$ $$\varphi'_x=1,\qquad \varphi'_y=1$$

$$\Delta=-\begin{vmatrix}0&1&1\\1&2&0\\1&0&2\end{vmatrix}=-(-2-2)=4>0.$$ Therefore, at point $P(1, 1)$, the function has a constrained minimum.

$z_{min}=1+1=2$

Answer: At point $P(1, 1)$, the function has a constrained minimum; $z_{min}=2.$

3. $z=2x+y$ subject to $x^2+y^2=1.$

Solution.

Let's form the Lagrange function:

$$L(x, y, \lambda)=2x+y+\lambda(x^2+y^2-1).$$

We have $$\frac{\partial L}{\partial x}=2+2x\lambda,\,\,\,\frac{\partial L}{\partial y}=1+2y\lambda.$$

System (1) takes the form $$\left\{\begin{array}{lcl}2+2x\lambda=0\\1+2y\lambda=0\\x^2+y^2-1=0\end{array}\right.$$

We solve the system:

$$\left\{\begin{array}{lcl}2+2x\lambda=0\\1+2y\lambda=0\\x^2+y^2-1=0\end{array}\right.\Rightarrow\left\{\begin{array}{lcl}x=-\frac{1}{\lambda}\\y=-\frac{1}{2\lambda}\\x^2+y^2-1=0\end{array}\right.\Rightarrow\left\{\begin{array}{lcl}x=-\frac{1}{\lambda}\\y=-\frac{1}{2\lambda}\\\frac{1}{\lambda^2}+\frac{1}{4\lambda^2}-1=0\end{array}\right.\Rightarrow$$

$$\Rightarrow\left\{\begin{array}{lcl}x=-\frac{1}{\lambda}\\y=-\frac{1}{2\lambda}\\{\lambda^2}=\frac{5}{4}\end{array}\right.\Rightarrow\left\{\begin{array}{lcl}\lambda=\pm\frac{\sqrt 5}{2}\\x=\mp\frac{2}{\sqrt 5}\\y=\mp\frac{1}{\sqrt 5}\end{array}\right.$$

We obtained two solutions of the system: for $\lambda_1=\frac{\sqrt 5}{2}$, $x_1=-\frac{2}{\sqrt 5}$, $y_1=-\frac{1}{\sqrt 5}$;

for $\lambda_2=-\frac{\sqrt 5}{2}$, $x_2=\frac{2}{\sqrt 5}$, $y_2=\frac{1}{\sqrt 5}$.

Next, we find the second derivatives: $$\frac{\partial^2 L}{\partial x^2}=\frac{\partial^2 L}{\partial y^2}=2\lambda,\,\,\,\frac{\partial^2 L}{\partial x\partial y}=0;$$ $$\varphi'_x=2x,\qquad \varphi'_y=2y$$

$$\Delta=-\begin{vmatrix}0&2x&2y\\2x&2\lambda&0\\2y&0&2\lambda\end{vmatrix}=-(-8\lambda y^2-8\lambda x^2).$$

$$\Delta(P_1(x_1, y_1))=-(-8\frac{\sqrt 5}{2} \frac{1}{5}-8\frac{\sqrt 5}{2} \frac{4}{5})=4\sqrt 5>0.$$

Therefore, at point $P_1\left(-\frac{2}{\sqrt{5}}, -\frac{1}{\sqrt{5}}\right)$, the function has a constrained minimum.

$$\Delta(P_2(x_2, y_2))=-(-8\frac{-\sqrt 5}{2} \frac{1}{5}-8\frac{-\sqrt 5}{2} \frac{4}{5})=-4\sqrt 5<0.$$

Therefore, at point $P_2\left(\frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right)$, the function has a constrained maximum.

$z_{min}=z(P_1)=-\frac{4}{\sqrt 5}-\frac{1}{\sqrt 5}=-\sqrt 5.$

$z_{max}=z(P_2)=\frac{4}{\sqrt 5}+\frac{1}{\sqrt 5}=\sqrt 5.$

Answer: At point $P_1\left(-\frac{2}{\sqrt{5}}, -\frac{1}{\sqrt{5}}\right)$, the function has a constrained minimum; $z_{min}=-\sqrt{5}$. At point $P_2\left(\frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right)$, the function has a constrained maximum; $z_{max}=\sqrt{5}$.

Tags: Investigation of functions, Necessary condition for an extremum, calculus, extremum, local extrema, mathematical analysis