# Finding absolute maximum and minimum values on circular bounded region

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!

• I think the photo didn't upload and try to use MathJax for math formatting. – Aaratrick Aug 9 '19 at 0:43
• @Aaratrick you're right, sorted all that, sorry! – Isidora Conic Aug 9 '19 at 1:42

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 $$\pm\frac8{3\sqrt3}$$.