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I have two functions $f=xy^2$ and $g=x^2+y^2$. When optimizing $xy^2$ on the circle $x^2+y^2=1$ I get 6 critical points but when I try to perform the second derivative test, it equals 0, meaning that the result is inconclusive. How can I find whether the critical points are in fact a maximum or minimum in this case?

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You can go this way. Since $x^2+y^2=1$ then $y^2=1-x^2$ and substituting in $f$ gives the one variable function

$$ f(x) = x(1-x^2). $$

Now you can use the derivative test to find max and min.

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    $\begingroup$ I think you need to optimize this equation using lagrange multipliers otherwise not all of the solutions can be found. $\endgroup$ – andrew749 Mar 18 '15 at 20:52
  • $\begingroup$ Since the tags say "calculus and optimization" so I gave you this alternative technique because you said the test you use was not conclusive. You can use Lagrange multiplier as a different approach. Anyways I gave you a technique that saves your time. $\endgroup$ – science Mar 18 '15 at 20:57
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    $\begingroup$ It is simply not the case that one must use Lagrange multipliers. I agree however that this solution is incomplete. This approach yields candidate points: $x=\pm 1/\sqrt{3}$, and the second derivative test confirms $+1/\sqrt{3}$ is the proper candidate. We must also consider $x=\pm 1$, since they are extremes, but they are easily eliminated since they require $y=0$. We arrive at two solutions: $(x,y)=(\sqrt{1/3},\pm\sqrt{2/3})$, optimal value $2/(3\sqrt{3})$. No Lagrange multipliers needed. +1 for @science! $\endgroup$ – Michael Grant Mar 19 '15 at 19:46
  • $\begingroup$ @MichaelGrant: Thank you for providing these details. I appreciate your comment. $\endgroup$ – science Mar 19 '15 at 20:17
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The best way to solve this problem is the method of Lagrange multipliers.Also, by the extreme value theorem,since f is defined on a closed and bounded subset of $R^2$, we know it must contain absolute extrema on the circle.

The method of Lagrange multipliers essentially considers the gradients of both the function f(x,y)=$xy^{2}$ and the level curve constraining function-in this case, the circle g(x,y)= $x^2 + y^2$ =1.(Strictly speaking, a circle isn't a function in the precise definition of the term.But since this is a basic calculus problem, we won't get into the weeds on this.) Since the direction of maximum change occurs along the gradients and f is constrained by g, we would expect these gradients to be parallel to each other. $\nabla {f(x,y}$ = $\lambda\nabla {g(x,y)}$ $\rightarrow$ $(f_x, f_y)$= $\lambda(g_x,g_y)$

So now let's compute each of the partial derivatives. Once we solve the resulting system for x and y, we plug in those values into f and find the extrema.

$f_x$ = $y^2$ $f_y$ = 2xy $g_x$ = 2x $g_y$ = 2y So $y^2$ = $\lambda 2x$ 2xy = $\lambda 2y$

Solving, we get x = $\lambda$ and y = $\pm \sqrt{2}x$.

So plugging into the constraining circle gives:

$x^2 + y^2$ =1 $\rightarrow$ $x^2 + (\pm\sqrt{2}x)^2$ =1 $\rightarrow$ $ 3x^2$ =1 $\rightarrow$ x = $\lambda$ = $\pm \frac {1}{3}$. So y = $\pm \frac {2}{3}$.

So now we have everything we need to solve the problem by plugging x and y into f(x,y). The level curves of the functions give a good indication of the geometry of the domain and it's local extrema:

enter image description here

The full graph in $R^3$ gives:

enter image description here

You can and should finish it now yourself. I only obtained 4 extrema-what were the other 2 you got? Unless I screwed up,of course, which I'm sure I'll hear about soon if so...............

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  • $\begingroup$ You're missing some square roots in there... $\endgroup$ – Michael Grant Mar 19 '15 at 19:44

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