4: Extrema Subject to Two Constraints

Here is Theorem [theorem:1] with $m=2$.

Theorem $\PageIndex1$

Nov 2, 2019, 8:44 PM

[theorem:3]

Suppose that $n>2.$ If  ${\bf X}_{0}$ is a local extreme point of $f$ subject to $g_1({\bf X})=g_2({\bf X})=0$ and

$$\label{eq:19} \left|\begin{array}{ccccccc} \displaystyle{\frac{\partial g_1({\bf X}_{0})}{\partial x_{r}}}& \displaystyle{\frac{\partial g_1({\bf X}_{0})}{\partial x_{s}}}\\\\ \displaystyle{\frac{\partial g_2({\bf X}_{0})}{\partial x_{r}}}& \displaystyle{\frac{\partial g_2({\bf X}_{0})}{\partial x_{s}}}\\ \end{array}\right|\ne0$$

for some $r$ and $s$ in $\{1,2,\dots,n\},$ then there are constants $\lambda$ and $\mu$ such that

$$\label{eq:20} \frac{\partial f({\bf X}_{0})}{\partial x_{i}}- \lambda\frac{\partial g_1({\bf X}_{0})}{\partial x_{i}}- \mu\frac{\partial g_2({\bf X}_{0})}{\partial x_{i}}=0,$$

$1\le i\le n$.

For notational convenience, let $r=1$ and $s=2$. Denote

$${\bf U}=(x_{3},x_{4},\dots x_{n})\text{ and } {\bf U}_{0}=(x_{30},x_{30},\dots x_{n0}). \nonumber$$

Since

$$\label{eq:21} \left|\begin{array}{ccccccc} \displaystyle{\frac{\partial g_1({\bf X}_{0})}{\partial x_1}}& \displaystyle{\frac{\partial g_1({\bf X}_{0})}{\partial x_2}}\\\\ \displaystyle{\frac{\partial g_2({\bf X}_{0})}{\partial x_1}}& \displaystyle{\frac{\partial g_2({\bf X}_{0})}{\partial x_2}}\\ \end{array}\right|\ne0,$$

the Implicit Function Theorem (Theorem 6.4.1) implies that there are unique continuously differentiable functions

$$h_1=h_1(x_{3},x_{4},\dots,x_{n})\text{ and } h_2=h_1(x_{3},x_{4},\dots,x_{n}), \nonumber$$

defined on a neighborhood $N\subset{\mathbb R}^{n-2}$ of ${\bf U}_{0},$ such that $(h_1({\bf U}),h_2({\bf U}),{\bf U})\in D$ for all ${\bf U}\in N$, $h_1({\bf U}_{0})=x_{10}$, $h_2({\bf U}_{0})=x_{20}$, and

$$\label{eq:22} g_1(h_1({\bf U}),h_2({\bf U}),{\bf U})= g_2(h_1({\bf U}),h_2({\bf U}),{\bf U})=0,\quad {\bf U}\in N.$$

From $\ref{eq:21}$, the system

$$\label{eq:23} \left[\begin{array}{ccccccc} \displaystyle{\frac{\partial g_1({\bf X}_{0})}{\partial x_1}}& \displaystyle{\frac{\partial g_1({\bf X}_{0})}{\partial x_2}}\\\\ \displaystyle{\frac{\partial g_2({\bf X}_{0})}{\partial x_1}}& \displaystyle{\frac{\partial g_2({\bf X}_{0})}{\partial x_2}}\\ \end{array}\right] \left[\begin{array}{ccccccc} \lambda\\\mu \end{array}\right]= \left[\begin{array}{ccccccc} f_{x_1}({\bf X}_{0})\\f_{x_2}({\bf X}_{0})\\ \end{array}\right]$$

has a unique solution (Theorem 6.1.13). This implies Equation $\ref{eq:20}$ with $i=1$ and $i=2$. If $3\le i\le n$, then differentiating Equation $\ref{eq:22}$ with respect to $x_{i}$ and recalling that $(h_1({\bf U}_{0}),h_2({\bf U}_{0}),{\bf U}_{0})={\bf X}_{0}$ yields

$$\frac{\partial g_1({\bf X}_{0})}{\partial x_{i}}+ \frac{\partial g_1({\bf X}_{0})}{\partial x_1} \frac{\partial h_1({\bf U}_{0})}{\partial x_{i}}+ \frac{\partial g_1({\bf X}_{0})}{\partial x_2} \frac{\partial h_2({\bf U}_{0})}{\partial x_{i}}=0 \nonumber$$

and

$$\frac{\partial g_2({\bf X}_{0})}{\partial x_{i}}+ \frac{\partial g_2({\bf X}_{0})}{\partial x_1} \frac{\partial h_1({\bf U}_{0})}{\partial x_{i}}+ \frac{\partial g_2({\bf X}_{0})}{\partial x_2} \frac{\partial h_2({\bf U}_{0})}{\partial x_{i}}=0. \nonumber$$

If ${\bf X}_{0}$ is a local extreme point of $f$ subject to $g_1({\bf X})=g_2({\bf X})=0$, then ${\bf U}_{0}$ is an unconstrained local extreme point of $f(h_1({\bf U}),h_2({\bf U}),{\bf U})$; therefore,

$$\frac{\partial f({\bf X}_{0})}{\partial x_{i}}+ \frac{\partial f({\bf X}_{0})}{\partial x_1} \frac{\partial h_1({\bf U}_{0})}{\partial x_{i}}+ \frac{\partial f({\bf X}_{0})}{\partial x_2} \frac{\partial h_2({\bf U}_{0})}{\partial x_{i}}=0. \nonumber$$

The last three equations imply that

$$\left|\begin{array}{ccccccc} \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_{i}}} & \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_1}}& \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_2}}\\\\ \displaystyle{\frac{\partial g_1({\bf X}_{0})}{\partial x_{i}}}& \displaystyle{\frac{\partial g_1({\bf X}_{0})}{\partial x_1}} & \displaystyle{\frac{\partial g_1({\bf X}_{0})}{\partial x_2}} \\\\ \displaystyle{\frac{\partial g_2({\bf X}_{0})}{\partial x_{i}}}& \displaystyle{\frac{\partial g_2({\bf X}_{0})}{\partial x_1}}& \displaystyle{\frac{\partial g_2({\bf X}_{0})}{\partial x_2}}\\ \end{array}\right|=0, \nonumber$$

$$\left|\begin{array}{ccccccc} \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_{i}}}& \displaystyle{\frac{\partial g_1({\bf X}_{0})}{\partial x_{i}}} & \displaystyle{\frac{\partial g_2({\bf X}_{0})}{\partial x_{i}}} \\\\ \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_1}} & \displaystyle{\frac{\partial g_1({\bf X}_{0})}{\partial x_1}} & \displaystyle{\frac{\partial g_2({\bf X}_{0})}{\partial x_1}} \\\\ \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_2}}& \displaystyle{\frac{\partial g_1({\bf X}_{0})}{\partial x_2}}& \displaystyle{\frac{\partial g_2({\bf X}_{0})}{\partial x_2}} \\\\ \end{array}\right|=0. \nonumber$$

Therefore, there are constants $c_1$, $c_2$, $c_{3}$, not all zero, such that

$$\label{eq:24} \left[\begin{array}{ccccccc} \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_{i}}} & \displaystyle{\frac{\partial g_1({\bf X}_{0})}{\partial x_{i}}} & \displaystyle{\frac{\partial g_2({\bf X}_{0})}{\partial x_{i}}} \\\\ \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_1}}& \displaystyle{\frac{\partial g_1({\bf X}_{0})}{\partial x_1}}& \displaystyle{\frac{\partial g_2({\bf X}_{0})}{\partial x_1}}\\\\ \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_2}}& \displaystyle{\frac{\partial g_1({\bf X}_{0})}{\partial x_2}} & \displaystyle{\frac{\partial g_2({\bf X}_{0})}{\partial x_2}}\\\\ \end{array}\right] \left[\begin{array}{ccccccc} c_1\\c_2\\c_{3} \end{array}\right]= \left[\begin{array}{ccccccc} 0\\0\\0 \end{array}\right].$$

If $c_1=0$, then

$$\left[\begin{array}{ccccccc} \displaystyle{\frac{\partial g_1({\bf X}_{0})}{\partial x_1}} & \displaystyle{\frac{\partial g_1({\bf X}_{0})}{\partial x_2}} \\\\ \displaystyle{\frac{\partial g_2({\bf X}_{0})}{\partial x_1}} & \displaystyle{\frac{\partial g_2({\bf X}_{0})}{\partial x_2}} \\ \end{array}\right] \left[\begin{array}{ccccccc} c_2\\c_{3} \end{array}\right]= \left[\begin{array}{ccccccc} 0\\0 \end{array}\right], \nonumber$$

so Equation $\ref{eq:19}$ implies that $c_2=c_{3}=0$; hence, we may assume that $c_1=1$ in a nontrivial solution of Equation $\ref{eq:24}$. Therefore,

$$\label{eq:25} \left[\begin{array}{ccccccc} \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_{i}}}& \displaystyle{\frac{\partial g_1({\bf X}_{0})}{\partial x_{i}}}& \displaystyle{\frac{\partial g_2({\bf X}_{0})}{\partial x_{i}}}\\\\ \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_1}}& \displaystyle{\frac{\partial g_1({\bf X}_{0})}{\partial x_1}} & \displaystyle{\frac{\partial g_2({\bf X}_{0})}{\partial x_1}} \\\\ \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_2}} & \displaystyle{\frac{\partial g_1({\bf X}_{0})}{\partial x_2}} & \displaystyle{\frac{\partial g_2({\bf X}_{0})}{\partial x_2}} \\\\ \end{array}\right] \left[\begin{array}{ccccccc} 1\\c_2\\c_{3} \end{array}\right]= \left[\begin{array}{ccccccc} 0\\0\\0 \end{array}\right],$$

which implies that

$$\left[\begin{array}{ccccccc} \displaystyle{\frac{\partial g_1({\bf X}_{0})}{\partial x_1}} & \displaystyle{\frac{\partial g_1({\bf X}_{0})}{\partial x_2}} \\\\ \displaystyle{\frac{\partial g_2({\bf X}_{0})}{\partial x_1}} & \displaystyle{\frac{\partial g_2({\bf X}_{0})}{\partial x_2}} \\ \end{array}\right] \left[\begin{array}{ccccccc} -c_2\\-c_{3} \end{array}\right]= \left[\begin{array}{ccccccc} f_{x_1}({\bf X}_{0})\\f_{x_2}({\bf X}_{0})\\ \end{array}\right]. \nonumber$$

Since Equation $\ref{eq:23}$ has only one solution, this implies that $c_2=-\lambda$ and $c_2=-\mu$, so Equation $\ref{eq:25}$ becomes

$$\left[\begin{array}{ccccccc} \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_{i}}} & \displaystyle{\frac{\partial g_1({\bf X}_{0})}{\partial x_{i}}}& \displaystyle{\frac{\partial g_2({\bf X}_{0})}{\partial x_{i}}}\\\\ \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_1}}& \displaystyle{\frac{\partial g_1({\bf X}_{0})}{\partial x_1}} & \displaystyle{\frac{\partial g_2({\bf X}_{0})}{\partial x_1}} \\\\ \displaystyle{\frac{\partial f({\bf X}_{0})}{\partial x_2}} & \displaystyle{\frac{\partial g_1({\bf X}_{0})}{\partial x_2}}& \displaystyle{\frac{\partial g_2({\bf X}_{0})}{\partial x_2}}\\\\ \end{array}\right] \left[\begin{array}{rcccccc} 1\\-\lambda\\-\mu \end{array}\right]= \left[\begin{array}{ccccccc} 0\\0\\0 \end{array}\right]. \nonumber$$

Computing the topmost entry of the vector on the left yields Equation $\ref{eq:20}$.

Example $\PageIndex1$

Nov 2, 2019, 8:47 PM

[example:7]

Minimize

$$f(x,y,z,w) = x^2+y^2+z^2+w^2 \nonumber$$

subject to

$$\label{eq:26} x+y+z+w = 10 \text{ and } x-y+z+3w = 6.$$

Solution

Let

$$L = \frac{x^2+y^2+z^2+w^2}2-\lambda(x+y+z+w)-\mu(x-y+z+3w); \nonumber$$

then

$$\begin{align*} L_x &= x-\lambda-\mu \\ L_y & = y-\lambda+\mu \\ L_z & =& z-\lambda-\mu \\ L_w & = w-\lambda-3\mu,\end{align*}$$

so

$$\label{eq:27} x_{0} = \lambda+\mu, \quad y_{0} = \lambda-\mu, \quad z_{0} = \lambda+\mu, \quad w_{0} = \lambda+3\mu.$$

This and Equation $\ref{eq:26}$ imply that

$$\begin{aligned} (\lambda+\mu)+(\lambda-\mu)+(\lambda+\mu) + (\lambda+3\mu) & =& 10 \\ (\lambda+\mu)-(\lambda-\mu)+(\lambda+\mu)+ (3\lambda+9\mu) & =& \phantom16.\end{aligned} \nonumber$$

Therefore,

$$\begin{aligned} 4\lambda + \phantom14\mu & =& 10 \\ 4\lambda + 12\mu & = &\phantom16,\end{aligned} \nonumber$$

so $\lambda=3$ and $\mu = -1/2$. Now Equation $\ref{eq:27}$ implies that

$$(x_{0},y_{0},z_{0},w_{0}) = \left(\frac{5}2,\frac{7}2,\frac{5}2 \frac{3}2\right). \nonumber$$

Since $f(x,y,z,w)$ is the square of the distance from $(x,y,z,w)$ to the origin, it attains a minimum value (but not a maximum value) subject to the constraints; hence the constrained minimum value is

$$f\left(\frac{5}2,\frac{7}2,\frac{5}2, \frac{3}2\right)=27. \nonumber$$

Example $\PageIndex1$

Nov 2, 2019, 8:49 PM

[example:8]

The distance between two curves in $\mathbb{R}^2$ is the minimum value of

$$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}, \nonumber$$

where $(x_1,y_1)$ is on one curve and $(x_2,y_2)$ is on the other. Find the distance between the ellipse

$$x^2+2y^2=1 \nonumber$$

and the line

$$\label{eq:28} x+y=4.$$

Solution

We must minimize

$$d^2=(x_1-x_2)^2 + (y_1-y_2)^2 \nonumber$$

subject to

$$x_1^2 + 2y_1^2 =1 \text{ and } x_2+y_2 = 4. \nonumber$$

Let

$$L = \frac{(x_1-x_2)^2 + (y_1-y_2)^2 - \lambda(x_1^2 + 2y_1^2)}2 -\mu(x_2+y_2); \nonumber$$

then

$$\begin{aligned} L_{x_1}&=&x_1-x_2-\lambda x_1\\ L_{y_1}&=&y_1-y_2-2\lambda y_1\\ L_{x_2}&=&x_2-x_1-\mu\\ L_{y_2}&=&y_2-y_1-\mu,\end{aligned} \nonumber$$

so

$$\begin{aligned} x_{10}-x_{20}&=&\lambda x_{10} \text{\; \quad (i)}\\ y_{10}-y_{20}&=&2\lambda y_{10}\text{\quad (ii)}\\ x_{20}-x_{10}&=&\mu\text{\quad \quad \;\;(iii)} \\ y_{20}-y_{10}&=&\mu.\text{\quad \quad \;\;(iv)}\end{aligned} \nonumber$$

From (i) and (iii), $\mu=-\lambda x_{10}$; from (ii) and (iv), $\mu=-2\lambda y_{10}$. Since the curves do not intersect, $\lambda\ne0$, so $x_{10}=2y_{10}$. Since $x_{10}^2+2y_{10}^2=1$ and $(x_{0},y_{0})$ is in the first quadrant,

$$\label{eq:29} (x_{10},y_{10})=\left(\frac2{\sqrt{6}},\frac1{\sqrt{6}}\right).$$

Now (iii), (iv), and Equation $\ref{eq:28}$ yield the simultaneous system

$$x_{20}-y_{20}=x_{10}-y_{10}=\frac1{\sqrt{6}},\quad x_{20}+y_{20}=4, \nonumber$$

so

$$(x_{20},y_{20}) = \left(2+\frac1{2\sqrt{6}}, 2-\frac1{2\sqrt{6}}\right). \nonumber$$

From this and Equation $\ref{eq:29}$, the distance between the curves is

$$\left[\left(2+\frac1{2\sqrt{6}} -\frac2{\sqrt{6}} \right)^2 + \left(2- \frac1{2\sqrt{6}} - \frac1{ \sqrt{6}}\right)^2\right]^{1/2} = \sqrt2 \left(2-\frac{3}{2\sqrt{6}}\right). \nonumber$$