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I need to find absolute maximum and minimum values of $$f(x,y)=4x^3+3y^2$$ on a unit disk $D={(x,y)|x^2+y^2\le 1}$

I thought about finding $f_x$ and $f_y$ first to find in the critical point is in $D$ or not. I think that it is because I found it to be $(0,0)$. From here we have to look at the boundary, am I correct? How do I proceed?

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Let $x = r\cos \theta, y = r\sin \theta \to f(r,\theta) = 4r^3\cos^3 \theta + 3r^2\sin^2 \theta=4r^3\cos^3\theta + 3r^2(1-\cos^2\theta)=3r^2+4r^3\cos^3\theta-3r^2\cos^2\theta$. Let $t = \cos \theta \to f(r,\theta) = f(r,t) = 3r^2+4r^3t^3-3r^2t^2$ with $0 \leq r \leq 1, -1 \leq t \leq 1$. Now taking partials we have: $f_r = 6r(1+2rt^3-t^2) = 0 = f_t = 6r^2t(2rt-1)$. Now if $rt \neq 0$, then: $1+2rt^3-t^2 = 0 = 2rt-1 \to 0=1 + t^2(2rt-1) = 1 + 0 = 1$, contradiction. Thus $rt = 0$. If $r = 0 = t \to x = 0 = y$. If $r = 0$ only ,then $x = r\cos \theta = 0 = r\sin \theta = y$. If $t = 0$ only, then $x = rt = 0, y = \pm r$. From this we conclude that: $f_{\text{localmin}} = 0, f_{\text{localmax}} = 3$, and considering the endpoints we evaluate $f(r,t)$ at $(r,t) = (0,\pm1),(1,\pm1)$, and have $f_{\text{absolutemin}} = -4, f_{\text{absolutemax}} = 4$.

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  • $\begingroup$ I already found the critical point to be $(0,0)$ Am I right? $\endgroup$ – Shrey Jul 14 '15 at 0:05
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    $\begingroup$ @Shrey yes. But notice how this answer creatively transformed your circular boundary (which was hard to work with) into a box (which should be much easier). $\endgroup$ – gt6989b Jul 14 '15 at 0:12
  • $\begingroup$ @KfSsOc Can you continue on with this approach? $\endgroup$ – Shrey Jul 14 '15 at 0:16
  • $\begingroup$ Is the minimum -4, and the maximum 4? $\endgroup$ – Shrey Jul 14 '15 at 0:21
  • $\begingroup$ Okay, So I was right? $\endgroup$ – Shrey Jul 14 '15 at 0:42
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You can either use Lagrange multipliers, or adopting your approach, on the boundary, note that $y^2 = 1-x^2$, so your function becomes $4x^3 + 3(1-x^2)$, can you find the extrema?

Compare the boundary extrema to the ones inside.

UPD for Lagrange multipliers, see some examples here: http://www.slimy.com/~steuard/teaching/tutorials/Lagrange.html#Examples

For your case, you minimize $f(x,y)$ subject to $g(x,y) = 0$, where $g(x,y) = x^2+y^2-1$. The multiplier $\lambda$ satisfies $\nabla f = \lambda \nabla g$, which for us is $(12x^2,6y) = \lambda (2x, 2y)$ and in addition we satisfy the constraint $g(x,y)=0$, so you have the system $$ \begin{split} 12x^2 &= 2\lambda x\\ 6y &= 2 \lambda y\\ x^2 + y^2 &= 1 \end{split} $$ From the second equation, clearly either $y=0$ (in which case $x = \pm 1$ from last equation) or $\lambda = 3$ (in which case $x=0$ from the first equation, and the last equation implies $y = \pm 1$). So the only solutions of the minimization problem are $(\pm 1, 0)$ and $(0, \pm 1)$. Evaluate to see which is better.

For the maximization problem, minimize $-f(x,y)$ instead.

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  • $\begingroup$ Would you mind adding more to your solution. I want to learn the lagrange multiplier method. $\endgroup$ – Shrey Jul 13 '15 at 23:59
  • $\begingroup$ @Shrey here is one half of an example for your case. $\endgroup$ – gt6989b Jul 14 '15 at 0:10
  • $\begingroup$ Let me work through it and I'll let you know what I get, can you check it after? $\endgroup$ – Shrey Jul 14 '15 at 0:11
  • $\begingroup$ @Shrey very late tonight, if I have time $\endgroup$ – gt6989b Jul 14 '15 at 0:12

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