Maximizing a function of two variables over a convex hull

Question

I saw this question in a mock exam:

Given positive real constants $a, b$ and $f(x,y)=(a x)^2+ abxy +(by)^2-a^2 x-b^2 y$, maximize $f$ over the interior and boundary of the convex hull of $(0,0), (1,0), (0,1)$.

What is the shortest way to answer this problem when sitting on a maths exam?

Answer 1

I don't think there are useful shortcuts here.

Check the inside

This is the usual optimization procedure where you are looking for stationary points satisfying $f_x(x,y) = 0 = f_y(x,y)$ and classifying them to get a maximum.

Check the boundary

Formally you could do Lagrange multipliers but since you are on one of 3 lines, 2 with easy form, this is easier by hand:

• along the $x$-axis, $y=0 \implies f(x,0) = (ax)^2 - a^2 x = a^2x(x-1)$ with max at $x=1/2$
• along the $y$-axis, $x=0 \implies f(0,y) = (by)^2 - b^2 y = b^2y(y-1)$ with max at $y=1/2$
• along $x=-y$ with $y=-x \implies f(x,-x) = a^2x(x-1) + b^2x(1-x)+ abx^2$

Can you finish?

Answer 2

Observe that $\operatorname{conv}\{(0, 0), (1, 0), (0, 1)\}$ is the triangle with edges (1.) $x = 0$ for $0 \leq y \leq 1,$ (2.) $y = 0$ for $0 \leq x \leq 1,$ and (3.) $y = -x$ for $0 \leq x \leq 1.$ Compute the critical points of $f(x, y)$ in the interior; then, compute the critical points on the boundary. (Explicitly, compute the critical points of (1.) $f(0, y)$ for $0 \leq y \leq 1,$ (2.) $f(x, 0)$ for $0 \leq x \leq 1,$ and (3.) $f(x, -x)$ for $0 \leq x \leq 1.$) By the Extreme Value Theorem, the maximum of the function is simply the largest among the values $f(C),$ where $C$ is among the above critical points.