
# 4: Extrema subject to Two Constraints


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

Theorem $$\PageIndex1$$

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

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

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

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

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

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

and

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

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,

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

The last three equations imply that

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

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

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

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

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

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

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

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

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

Example $$\PageIndex1$$

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}

Therefore,

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

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

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

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

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

Example $$\PageIndex1$$

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

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

Let

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

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}

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}

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

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

so

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

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

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