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 $P\ne P_0$ 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 $$L(x_1,x_2,...,x_n,{\lambda }_1,...,{\lambda }_m)=f(x_1,...,x_m)+\sum _{k=1}^m{\lambda }_k{\phi }_k(x_1,...,x_n);$$

${\lambda }_k(k=1,2,...,m)$ are called Lagrange multipliers.

The necessary conditions for a constrained extremum are expressed by a system of $n+m$ equations$$\begin{gathered} \frac{\partial L(P)}{\partial x_i}=0(i=1,2,...,n) \\ {\phi }_k(P)=0(k=1,2,...,m)(1) \end{gathered}$$ 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 $d^{2}L(x_1^0,,...,x_n^0,,{\lambda }_1^0,,...,,{\lambda }_m^0,d{x}_1,,...,d{x}_n)$ for each set of values $x_1^0,,...,x_n^0,,{\lambda }_1^0,,...,,{\lambda }_m^0$ obtained from (1) under the condition that $d{x}_1,,d{x}_2,,...,d{x}_n$ satisfy the equations$$\sum _{j=1}^n\frac{\partial {\phi }_k(x_1^0,...,x_n^0)}{\partial x_j}d{x}_j=0(k=1,2,...,m)(2)$$ where $d{x}_1^2+d{x}_2^2+...+d{x}_n^2\ne 0.$ 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 $d{x}_1,...,d{x}_n$ satisfying conditions (2) and not all equal to zero simultaneously, the inequality $d^{2}L(x_1^0,,...,x_n^0,,{\lambda }_1^0,,...,,{\lambda }_m^0,d{x}_1,,...,d{x}_n)<0$ holds, and it has a constrained minimum if under these conditions $d^{2}L(x_1^0,,...,x_n^0,,{\lambda }_1^0,,...,,{\lambda }_m^0,d{x}_1,,...,d{x}_n)>0.$

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

System (1) consists of three equations:

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

Let $P_0(x_0,y_0),,{\lambda }_0$ be any solution of this system, and $$\mathrm{\Delta }=-\left|\begin{array}{ccc}0& {\phi }_x^{\prime }(P_0)& {\phi }_y^{\prime }(P_0)\\ {\phi }_x^{\prime }(P_0)& L_{xx}^{″}(P_0,{\lambda }_0)& L_{xy}^{″}(P_0,{\lambda }_0)\\ {\phi }_y^{\prime }(P_0)& L_{xy}^{″}(P_0,{\lambda }_0)& L_{yy}^{″}(P_0,{\lambda }_0)\end{array}\right|.$$

If $\mathrm{\Delta }<0$, then the function $z=f(x,y)$ has a constrained maximum at point $P_0(x_0,y_0)$, and if $\mathrm{\Delta }>0$, 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: $$\{\begin{array}{l}2x-y+1+\lambda =0\\ 2y-x+1+\lambda =0\\ x+y+3=0\end{array}$$

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

$$\mathrm{\Delta }=\left|\begin{array}{ccc}2& -1& 1\\ -1& 2& 1\\ 1& 1& 0\end{array}\right|=-1-1-2-2=-6;$$ $${\mathrm{\Delta }}_1=\left|\begin{array}{ccc}-1& -1& 1\\ -1& 2& 1\\ -3& 1& 0\end{array}\right|=-1+3+6+1=9;$$ $${\mathrm{\Delta }}_2=\left|\begin{array}{ccc}2& -1& 1\\ -1& -1& 1\\ 1& -3& 0\end{array}\right|=3-1+1+6=9;$$ $${\mathrm{\Delta }}_3=\left|\begin{array}{ccc}2& -1& -1\\ -1& 2& -1\\ 1& 1& -3\end{array}\right|=-12+1+1+2+2+3=-3.$$ From here, we find: $$x=\frac{{\mathrm{\Delta }}_1}{\mathrm{\Delta }}=\frac{9}{-6}=-1,5;y=\frac{{\mathrm{\Delta }}_2}{\mathrm{\Delta }}=\frac{9}{-6}=-1,5;\lambda =\frac{{\mathrm{\Delta }}_3}{\mathrm{\Delta }}=\frac{-3}{-6}=0,5.$$

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

$$\mathrm{\Delta }=-\left|\begin{array}{ccc}0& 1& 1\\ 1& 2& -1\\ 1& -1& 2\end{array}\right|=-(-1-1-2-2)=6>0.$$ 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: $$\{\begin{array}{l}-\frac{1}{x^2}+\lambda =0\\ -\frac{1}{y^2}+\lambda =0\\ x+y-2=0\end{array}$$

Let's solve the system:

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

$$\Rightarrow \{\begin{array}{l}\lambda =\frac{1}{x^2}\\ 4-4y+y^2=y^2\\ x=2-y\end{array}\Rightarrow \{\begin{array}{l}\lambda =1\\ y=1\\ x=1\end{array}$$

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

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

$$\mathrm{\Delta }=-\left|\begin{array}{ccc}0& 1& 1\\ 1& 2& 0\\ 1& 0& 2\end{array}\right|=-(-2-2)=4>0.$$ 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 $$\{\begin{array}{l}2+2x\lambda =0\\ 1+2y\lambda =0\\ x^2+y^2-1=0\end{array}$$

We solve the system:

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

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

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: $$\frac{{\partial }^{2}L}{\partial x^2}=\frac{{\partial }^{2}L}{\partial y^2}=2\lambda ,\frac{{\partial }^{2}L}{\partial x\partial y}=0;$$ $${\phi }_x^{\prime }=2x,{\phi }_y^{\prime }=2y$$

$$\mathrm{\Delta }=-\left|\begin{array}{ccc}0& 2x& 2y\\ 2x& 2\lambda & 0\\ 2y& 0& 2\lambda \end{array}\right|=-(-8\lambda y^2-8\lambda x^2).$$

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

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

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

Therefore, at point $P_2(\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}})$, 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 $P_1(-\frac{2}{\sqrt{5}},-\frac{1}{\sqrt{5}})$, the function has a constrained minimum; $z_{min}=-\sqrt{5}$. At point $P_2(\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}})$, 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