Question

Find the absolute maximum and absolute minimum of the function f(x,y)=2x3+y4f(x,y)=2x3+y4 on the region {(x,y)|x2+y2≤64}{(x,y)|x2+y2≤64} Ignore unneeded answer blanks, and list points lexicographically.

Answer 1

Check where the first-order partial derivatives vanish to find any critical points within the given region:

$f(x,y)=2x^3+y^4\implies\begin{cases}f_x=6x^2=0\\f_y=4y^3=0\end{cases}\implies(x,y)=(0,0)$

The Hessian for this function is

$\mathbf H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}12x&0\\0&12y^2\end{bmatrix}$

with $\det\mathbf H(0,0)=0$, so unfortunately the second partial derivative test fails. However, if we take $x=0$ we see that $f(x,y)>0$ for different values of $y$; if we take $y=0$ we see $f(x,y)$ takes on both positive and negative values. This indicates (0, 0) is neither the site of an extremum nor a saddle point.

Now check for points along the boundary. We can parameterize the boundary by

$(x,y)=(8\cos t,8\sin t)$

with $0\le t\le2\pi$. This turns $f(x,y)$ into a univariate function $F(t)$:

$F(t)=f(8\cos t,8\sin t)=2^{10}\cos^3t+2^{12}\sin^4t$

$\implies F'(t)=3\cdot2^{10}\cos^2t(-\sin t)+2^{14}\sin^3t\cos t=2^{10}\sin t\cos t(16\sin^2t-3\cos t)$

$F'(t)=0\implies\begin{matrix}\sin t=0\implies t=0,\,t=\pi\\\\\cos t=0\implies t=\dfrac\pi2,\,t=\dfrac{3\pi}2\\\\16\sin^2t-3\cos t=0\implies t=2\tan^{-1}\sqrt{\dfrac{\sqrt{1033}-32}3},\,t=2\pi-2\tan^{-1}\sqrt{\dfrac{\sqrt{1033}-32}3}\end{matrix}$

At these critical points, we get

$F(0)=1024$

$F\left(2\tan^{-1}\sqrt{\dfrac{\sqrt{1033}-32}3}\right)\approx893$

$F\left(\dfrac\pi2\right)=4096$

$F(\pi)=-1024$

$F\left(\dfrac{3\pi}2\right)=4096$

$F\left(2\pi-2\tan^{-1}\sqrt{\dfrac{\sqrt{1033}-32}3}\right)\approx893$

We only care about 3 of these results.

$t=\dfrac\pi2\implies(x,y)=(0,8)$

$t=\pi\implies(x,y)=(-8,0)$

$t=\dfrac{3\pi}2\implies(x,y)=(0,-8)$

So to recap, we found that $f(x,y)$ attains

• a maximum value of 4096 at the points (0, 8) and (0, -8), and
• a minimum value of -1024 at the point (-8, 0).