Extrema of $f(x,y)=e^{x(y+1)}$ for $\sqrt{x^2+y^2} \le 1$

Question

How can I determine the extrema (minima and maxima) of $f: D \to \mathbb{R}$

$$f(x,y)=e^{x(y+1)}$$

for $D=\{(x,y) \in \mathbb{R}^2:\sqrt{x^2+y^2} \le 1\}?$

The solution should be $\Big(\frac{\sqrt{3}}{2},\frac{1}{2}\Big)$ for the maximum and $\Big(-\frac{\sqrt{3}}{2},\frac{1}{2}\Big)$ for the minimum.

Answer 1

HINT

1. Solve $f_x = 0 = f_y$ to find the local extrema, you are only interested in the ones inside $D$. You can use the 2nd derivative test to classify them.
2. Use Lagrange multipliers to solve the boundary problem looking for extreme values on $D$ itself.

Answer 2

This is a different way to do it than what you might be expected to do:

To maximize $e^{x(y+1)}$ we just have to maximize $x(y+1)$. Our domain is a a disk of radius $1$ centered at the origin; however, we can tell that the maximum will occur on the border rather than inside. Why? Let $O$ be the origin. For any point $P(x, y)$ inside the disk, if we continue along the ray OP until it hits the border, we will have both a larger $x$ and a larger $y$, so a larger $x(y+1)$. Thus we know that $x^2+y^2=1$, rather than $\le 1$.

Thus $y = \pm \sqrt {1-x^2}$, and we want to maximize $x(y+1)$ which is $x(\pm \sqrt {1-x^2}+1)$, which can be done with Calculus 1.