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

Take partial derivatives and set them equal to 0:

[tex]\nabla F(x,y)=\langle2x+4,2y-4\rangle=\mathbf 0[/tex]

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

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

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

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

[tex]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)[/tex]

with critical points at

[tex]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[/tex]

At these points, we get

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

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