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.

Answers

The answer maybbe c

8 2/3 is the answer

step-by-step explanation: just use a calculator

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), anda minimum value of -1024 at the point (-8, 0).


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