A question on Lagrange multipliers The state of Megalomania occupies the region $x^4 + y^4 \leq 30,000.$ The altitude at the point $(x,y)$ is $\frac{1}{8}xy+200x$ meters above sea level. Where are the highest and lowest points in the state?
I tried using the standard method of finding all the critical points inside the region and using the method of Lagrange multipliers on the boundary of the region to find the extrema. But I am not getting the right answers given at the back of the book. In fact, I am not even getting an answer.
 A: Your general approach seems to be correct. First we find any critical points inside the region: We define $f(x,y)=\frac{1}{8}xy+200x$. We calculate the partial derivatives
$$\frac{\partial f}{\partial x}=\frac{1}{8}y+200$$
$$\frac{\partial f}{\partial y}=\frac{1}{8}x$$
We set both of these equal to $0$ to find our only critical point is $(0,-25)$. But we can easily see this is not a local minimum or maximum, as moving along the line $y=x$ shows that, in any neighbourhood of our critical point, $f$ can be both greater and less than $f(0,-25)$.
So we turn our attention to the boundary.
We will use the method of Lagrange Multipliers, with $f$ as above and $g(x,y)=x^4+y^4$. We introduce the multiplier $\lambda$ and solve
$$\nabla(f-\lambda g)=0$$
which gives rise to the equations
$$\frac{1}{8}y+200-4\lambda x^3=0$$
$$\frac{1}{8}x-4\lambda y^3=0$$
We solve the second equation for $\lambda$ to give 
$$\lambda=\frac{1}{32}\frac{x}{y^3}$$
We substitute into the 1st equation to give
$$\frac{1}{8}y+200-\frac{1}{8}\frac{x^4}{y^3}=0$$
Solving this for $x^4$ gives
$$x^4=y^4+1600y^3$$
We substitute this into our condition on $g$ to give
$$2y^4+1600y^3=30000$$
This has 2 real solutions which we will call $y_1,y_2$, with $y1\approx -800$ and $y_2\approx 2.65$. We see that only $y_2$ is valid in our region.
So we find the real $x$ that correspond to $y2$. These are $x=\pm \left(y_2^4+1600y_2^3\right)^\frac{1}{4}$
These correspond to roughly $(-13.12,2.65),(13.12,2.65)$. As our boundary is a compact set, $f$ will attain its maximum and minimum values on the boundary, and in particular these will be critical points. By evaluating $f$ at the two points we’ve found, we see that $(—13.12,2.65)$ is the global minimum, and $(13.12,2.65)$ is the global maximum. So the point in the region of highest altitude is $(13.12,2.65)$, and the point of lowest altitude is $(-13.12,2.65)$.
