Not sure how to approach this system of exponential equations

Question

This is the problem I'm trying to solve: let $g(x,y)=e^{-3(x-1)^2}+e^{-3y^2}$ be the height of a mountain at a given $(x,y)$ point. Find max heights within $\frac{(x-1)^2}4+\frac{y^2}4 \le 1$

So what I first thought is Lagrange multipliers. So I got the gradients for both the surface and the restriction (which I called $f$):

$\nabla g=<-6e^{-3(x-1)^2}(x-1) ; -6ye^{-3y^2}>$

$\nabla f=<\frac{x-1}2; \frac y 2>$

So, given that $\nabla g = \lambda \nabla f$ my system of equations is: \begin{cases} -6e^{-3(x-1)^2}(x-1) = \lambda \frac{x-1}2 \\ -6ye^{-3y^2} = \lambda \frac y 2 \\ \frac{(x-1)^2}4+\frac{y^2}4 = 1 \end{cases}

If I set up my information the right way, I'm now completely stuck with this system of equations and how to solve it...

Answer 1

If we divide $x-1$ out of the first equation and $y$ out of the second, and then multiply the two equations, we get $(x-1)^2 + y^2$ in an exponent. From the third equaiton, we know $(x-1)^2 + y^2 = 4$. So this will allow us to solve for $\lambda$.

Answer 2

The right approach is always to try to intuit the problem... and to start using analytic methods when you already know the solution. At least in cases when intuition works.

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The plot of the function $$e^{-3(x-1)^2}+e^{-3y^2}$$ (by Alpha) is

By intuition, along the circle $$(x-1)^2+y^2=4$$ (of radius $2$) the extrema are at $(1,-2)$, $(3,0)$, $(1,2)$, and $(-1,0)$; and at four other places somewhere in between. Within the circle the absolute maximum is at $(1,0)$.

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Consider the extrema along the circle parametrized as $$(2\cos(\theta)+1, 2\sin(\theta)).$$ We have to find the extrema of $$e^{-12\cos^2(\theta)}+e^{-6\sin^2(\theta)}.$$

Take the derivative with respect to $\theta$ and set it equal to zero:

$$12\cos(\theta)\sin(\theta)\left(2e^{-12\cos^2(\theta)}-e^{-6\sin^2(\theta)}\right)=0.$$

This will hold if either $\cos(\theta)$ or $\sin(\theta)=0$, that is when $$\theta=0,\frac{\pi}2,\pi, \frac{3\pi}2$$ as intuition has shown. In order to find the "in between" places, we have to solve the following equation

$$e^{-12\ln(2)\cos^2(\theta)}=e^{-6\sin^2(\theta)}.$$

Taking the natural logarithm of both sides, and then taking the square roots of the samewe get $$\mid \ \cos(\theta)\mid=\sqrt{\frac1{{\ln(4)}}}\mid\sin(\theta)\mid.$$

Solving this equation will give the further four extrema as shown below

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As far as the absolute maximum. Examine the function as a function of $x$ with fixed $y$s and as a function of $y$ with fixed $x$s. The shape is always like this

if $y$ is fixed, and like this

if $x$ is fixed...