Question

F(x, y) = x2 + y2 + 4x − 4y, x2 + y2 ≤ 81 find the extreme values of f on the region described by the inequality.

Answer 1

Take partial derivatives and set them equal to 0:

$\nabla F(x,y)=\langle2x+4,2y-4\rangle=\mathbf 0$

We find one critical point within the boundary of the disk at $(x,y)=(-2,2)$. The Hessian matrix for this function is

$\mathbf H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2&0\\0&2\end{bmatrix}$

which is positive definite, and incidentally independent of $x$ and $y$, so $F(x,y)$ attains a minimum $F(-2,2)=-8$.

Meanwhile, we can parameterize the boundary by $\mathbf r(t)=\langle9\cos t,9\sin t\rangle$ with $0\le t\le2\pi$, which gives

$F(x,y)=F(x(t),y(t))=f(t)=81\cos^2t+81\sin^2t+36\cos t-36\sin t=81+36(\cos t-\sin t)$

with critical points at

$f'(t)=0\implies 36(-\sin t-\cos t)=0\implies\sin t+\cos t=0\implies t=\dfrac{3\pi}4,\dfrac{7\pi}4$

At these points, we get

$f\left(\dfrac{3\pi}4\right)=81-36\sqrt2\approx30.0883$
$f\left(\dfrac{7\pi}4\right)=81+36\sqrt2\approx131.9117$

so we attain a maximum only when $t=\dfrac{7\pi}4$, which translates to $(x,y)=\left(\dfrac9{\sqrt2},-\dfrac9{\sqrt2}\right)$.