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Find the absolute maximum and minimum values of $f(x,y)=4x^2y$ on the set $S=\{(x,y):x^2+y^2\le1\}$.

I am confused as to how to check the boundary of the circular region. I tried subtracting the formulas, i.e. $z = 4(x^2)y - 9 = x^2 + y^2$ and got the critical points of $(\frac{1}{2\sqrt 2} , \frac{1}{4})$ and $(-\frac{1}{2\sqrt 2} , \frac{1}{4})$ but this seems to be incorrect?

Any help would be much appreciated, preferably building on the method I used (unless it is a completely wrong approach).

Thank you so much!

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  • $\begingroup$ I think the photo didn't upload and try to use MathJax for math formatting. $\endgroup$ – Aaratrick Aug 9 '19 at 0:43
  • $\begingroup$ @Aaratrick you're right, sorted all that, sorry! $\endgroup$ – Isidora Conic Aug 9 '19 at 1:42
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Your method is incorrect. The boundary can be parametrised as $x=\cos t,y=\sin t$, so we have $f(x,y)=4\cos^2t\sin t=4(\sin t-\sin^3t)$. By considering this as a polynomial in $\sin t$ over $[-1,1]$, we see that $f$ attains its maximum and minimum of $\pm\frac8{3\sqrt3}$ when $\sin t=\pm\frac1{\sqrt3}$ respectively.

We also need to check the interior of the disc. The gradient of $f$ is $(8xy,4x^2)$ and this is only zero at $(0,0)$, but then $f(x,y)=0$ which is not an extremum. Thus the absolute extrema on $S$ are $\pm\frac8{3\sqrt3}$.

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