## Lagrange Multipliers

1. So, last time we considered finding extrema for some function ${f\colon{\mathbb R}^{n}\rightarrow{\mathbb R}}$, but what if we constrain our focus to some surface ${g\colon{\mathbb R}^{n}\rightarrow{\mathbb R}}$? For example, the unit circle would have
(1)$\displaystyle g(x,y) = x^{2}+y^{2} - 1=0$
How do we find extrema for ${f(x,y)=xy}$ on the unit circle?

2. What can we do? First we can consider the level curves ${f(x,y)=c}$. These are precisely the curves ${y=c/x}$. The gradient vector for ${f}$ is precisely
(2)$\displaystyle \nabla f=\langle y,x\rangle.$
This points in the direction of increasing values of ${f}$.

3. We want to consider the situation when ${\nabla f=\lambda\nabla g}$, i.e., when the gradient of ${f}$ is precisely a scaled tangent of ${g}$. Why? Because we want the gradient of ${f}$ to point in the direction of a tangent to our surface. We draw the circle, the gradient vector ${\nabla f}$ in red, and ${\nabla g}$ in blue. Remember, the red vectors point in the direction of increasing values of ${f}$, and we restrict our movement along the circle:

Observe when the red and blue vectors are perpendicular, ${f=0}$. But when they overlap as a purple vector or point in completely opposite direction, what happens?

This happens when
(3)$\displaystyle \langle y,x\rangle = \lambda\langle 2x,2y\rangle$
or equivalently
(4)$\displaystyle y=2\lambda x,\quad\mbox{and}\quad x=2\lambda y.$
Solving for ${2\lambda}$, we find
(5)$\displaystyle 2\lambda = \frac{y}{x} = \frac{x}{y}$
which implies
(6)$\displaystyle x^{2}=y^{2}.$
But we’re not quite done!

4. We must remain on the circle, so we also must demand that ${x^{2}+y^{2}=1}$. This equivalently implies
(7)$\displaystyle x^{2}=\frac{1}{2}\implies x=\pm\frac{\sqrt{2}}{2}.$

5. Is this optimal? Lets try approaching the problem differently. We are working on the circle, which is the parametric curve
(8)$\displaystyle x(t) = \cos(t),\quad\mbox{and}\quad y(t) = \sin(t).$
Thus the function we are optimizing becomes
(9)\displaystyle \begin{aligned} f(t) &= f\bigl(x(t),y(t)\bigr)\\ &=\cos(t)\sin(t) \end{aligned}
We see
(10)$\displaystyle f'(t) = -\sin^{2}(t)+\cos^{2}(t) = 0.$
We need to solve for ${t}$, to do so we rearrange terms
(11)$\displaystyle \sin^{2}(t)=\cos^{2}(t)$
and divide through by ${\cos^{2}(t)}$, getting
(12)$\displaystyle \tan^{2}(t) = 1.$
But this implies ${t=(2n+1)\pi/4}$ where ${n=0}$, ${1}$, ${2}$, or ${3}$. Look: that’s precisely describing ${(x,y)=(\pm1/\sqrt{2},\pm1/\sqrt{2})}$.

6. Lagrange Multipliers, Constraints. One way to consider this situation “Optimize ${f}$ subject to the constraint ${g=0}$” is to say Okay, so suppose ${g=0}$, then wouldn’t we have
(13)$\displaystyle f + \lambda g\approx f?$
The ${g}$ term vanishes anyways, so intuitively it seems “equal-ish”.

Example 1 (Minimizing Surface Area). Find the dimensions of the cylinder with smallest surface area whose volume is fixed at ${16\pi}$.

Solution: The outline takes several steps, namely, (1) construct the function, (2) take the derivatives, (3) solve.

Step One: Construct the Functions. We first write
(14)$\displaystyle A(r,h) = \pi r^{2} + \pi r^{2} + 2\pi rh$
for the surface area, and
(15)$\displaystyle V(r,h) = \pi r^{2}h$
describes the area. The constraint is
(16)$\displaystyle C(r,h) = V(r,h) - 16\pi.$
Thus we construct the function
(17)$\displaystyle F(r,h) = A(r,h) - \lambda C(r,h).$
This concludes the first step.

Step Two: Take the Derivatives. We find
(18)$\displaystyle \nabla F(r,h) = \langle \partial_{r}A(r,h) - \lambda\partial_{r}C(r,h), \partial_{h}A(r,h)-\lambda\partial_{h}C(r,h)\rangle = 0.$
Observe
(19)$\displaystyle \partial_{r}A(r,h)=4\pi r+2\pi h,\quad\mbox{and}\quad \partial_{r}C(r,h)=2\pi rh.$
We also have
(20)$\displaystyle \partial_{h}A(r,h)=2r\pi,\quad\mbox{and}\quad \partial_{h}C(r,h)=\pi r^{2}.$
The derivative with respect to the Lagrange multiplier gives us
(21)$\displaystyle \partial_{\lambda}F(r,h) = C(r,h) = 0.$
So we can set up our equations as
(22)\displaystyle \begin{aligned} (\partial_{r}F&=0)&\quad &\quad &2\pi r+2\pi h&=\lambda 2\pi rh\\ (\partial_{h}F&=0)&\quad &\quad &2\pi r &=\lambda\pi r^{2}\\ (\partial_{\lambda}F&=0)&\quad &\quad &\pi r^{2}h-16\pi&=0. \end{aligned}
That concludes our second step.

Step Three: Solve. We see immediately from the ${\partial_{h}F}$ equation that
(23)$\displaystyle 2\pi r = \lambda \pi r^{2}\implies \lambda=\frac{2}{r}.$
We plug this into the ${\partial_{r}F}$ equation, we get
(24)\displaystyle \begin{aligned} 2\pi r+2\pi h &= \lambda2\pi rh\\ &=\left(\frac{2}{r}\right)2\pi rh\\ &=4\pi h \end{aligned}
and subtracting ${2\pi h}$ from both sides yields
(25)$\displaystyle 2\pi r=2\pi h\implies r=h.$
Now we use the constraint
(26)$\displaystyle \pi r^{2}h = 16\pi$
substituting ${r=h}$ we get
(27)$\displaystyle \pi h^{3} = 16\pi\implies h = \sqrt[3]{16}.$
Thus when we take ${r=h=\sqrt[3]{16}}$, we minimize the surface area.

Exercise 1. Find the extrema of ${f(x,y,z)=xy+yz+zx}$ subject to the constraint ${g(x,y,z)=x^{2}+y^{2}+z^{2}-1}$.

Exercise 2. Find the extrema of ${f(x,y)=\exp(-xy)}$ subject to the constraint ${g(x,y)=x^{2}+y^{2}-1}$.