Using the Lagrange multipliers to solve an optimization question

Question

I'm having trouble with a question given to us as an extra exercise before our exams. The question states that there is a town that lives near a hill. The height at position $(x,y)$ can be described by the function:

$$H(x,y)=100e^{-\frac{x^2}{8}-\frac{y^2}{2}+x}$$

where $(x,y)$ are coordinates relative to the center of the city in meters. The major of the city wants to take an aerial photograph using a drone which flies in a circular path of equation $x^2+y^2=4$ and stays at a constant height. What is the minimum height the drone has to fly so that it never hits the hill.

So far i managed to setup the equation

$$L=100e^{-\frac{x^2}{8}-\frac{y^2}{2}+x} +\lambda(x^2+y^2-4)$$

I differentiated with respect to $x, y,$ and $\lambda$ and got the three equations:

$$(100-25x)e^{-\frac{x^2}{8}-\frac{y^2}{2}+x}+2x\lambda =0$$ $$-100ye^{-\frac{x^2}{8}-\frac{y^2}{2}+x}+2y\lambda = 0$$ $$x^2+y^2-4=0$$

I know that i need to solve these to find my $x$ and $y$ value that will give a minimum $H$, but I don't see how I would do this. The exam is non calculator. Have I made a mistake in my work?

Thanks in advance for any help.

Edit: I mistakenly wrote $\frac{x^2}{2}$ instead of $\frac{x^2}{8}$

Answer 1

i would write $$10e^{-\frac{1}{2}(x^2+y^2-x)}=100e^{-\frac{1}{2}(4-x)}$$

Answer 2

Since $H=100e^{x-2}$, you just need to maximise $x$. Then $x=2$ so $H=100$.

Answer 3

Move lambda terms to the right side and divide the first equation by the second to find $x=\frac43$.

From the third equation you find $y=\pm \sqrt{\frac{20}{9}}$.