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Find maximum and minimum values of $f(x,y,z)=x^2yz$ on the region $x^2+y^2\leq1,$ $0\leq z\leq1.$

First, I get $\nabla f= (2xyz, x^{2}z, x^{2}y) = (0,0,0) \implies x = y = z = 0$, so the critical point is $(0,0,0)$, and $f(0,0,0)=0$. There is no singular point, so I just need to consider the boundary points. From $x^{2}+y^{2}=1 \implies x^{2}=1-y^{2} \implies f(x,y,z)=(1-y^{2})yz = yz-y^{3}z = g$, then I get $\nabla g = (z-3zy^{2}, y-y^{3}) = (0,0)$ → critical point of $g$ are $(-1,0),(1,0),(0,0)$,then I insert $y = \pm 1, 0$ into $x^{2}=1-y^{2}$ → $x = 0 , \pm 1$ → the boundary point of $f$ are $(±1,0,0),(0,±1,0)$, and all the $4$ points lead $f=0$. So the max and min of $f$ are $0$.

Of course my idea is wrong, but I don't know where I did anything wrong, does anyone could help me, thanks a lot.

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  • $\begingroup$ The answer given is correct. You do restrict to the level set $x^2 = 1 - y^2$ and the problem reduces to two variables. Then, you want to look at the boundary of this problem, which is $z=0$ and $z=1$. This will reduce to maximising $y - y^3$ and this yields $1 - 3y^2 = 0$ and maximizing: $y = 1/\sqrt{3}$. So, $x = \sqrt{2/3}$ and $x^2\cdot y \cdot z = (2/3)\cdot 1/\sqrt{3} \cdot 1 = \frac{2}{3^{3/2}}$ if you want to maximize. Alternatively, taking $y = -1/\sqrt{3}$ and $z=1$, we get that the minimum is $\frac{-2}{3^{3/2}}$. $\endgroup$ – Chris K Mar 23 '14 at 2:51
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    $\begingroup$ $(2xyz, x^{2}z, x^{2}y) = (0,0,0)$ doesn’t imply that $x = y = z = 0$. $\endgroup$ – Steve Kass Mar 23 '14 at 2:56
  • $\begingroup$ Good point Steve. The OP should just state that since either $x=0$ or $y=z=0$, we have that $f(x,y,z) = 0$ for these cases. Then he can move on. $\endgroup$ – Chris K Mar 23 '14 at 3:00
  • $\begingroup$ I don't understand why the boundary points are on z=0 and z=1. If we draw the graph of the restriction, it should be a solid cylinder along z-axis, with height of 1 and radius of 1, so the boundary point could also be on the side face, not just the top and the bottom. $\endgroup$ – user133140 Mar 23 '14 at 3:25
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The answer given is correct. You do restrict to the level set $x^2 = 1 - y^2$ and the problem reduces to two variables. Then, you want to look at the boundary of this problem, which is $z=0$ and $z=1$. This will reduce to maximising $y - y^3$ and this yields $1 - 3y^2 = 0$ and maximizing: $y = \frac{1}{\sqrt{3}}$. So, $x = \sqrt{\frac{2}{3}}$ and $x^2\cdot y \cdot z = \frac{2}{3}\cdot \frac{1}{\sqrt{3}} \cdot 1 = \frac{2}{3^{3/2}}$ if you want to maximize. Alternatively, taking $y = -\frac{1}{\sqrt{3}}$ and $z=1$, we get that the minimum is $\frac{-2}{3^{3/2}}$. This is the tried and true approach for solving this problem. Of course, shortcuts are useful, but largely defeat the purpose of the exercise.

A note on how I think about this:

I like to think of this as we have a $3$-manifold with boundary and we take the gradient and set to zero. Then we restrict the function to the boundary, which is a $2$-manifold with boundary. Again, we take the gradient and set to zero; then, we restrict the function to the boundary, which is a $1$-manifold (in this case, it is without boundary. Why?). Then, if a $1$-manifold with boundary, we take the gradient again and reduce it to a $0$-manifold and check the discrete points left over. But this is my way of thinking about it; if you haven't discussed manifolds, then you can ignore this.

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  • $\begingroup$ I don't understand why the boundary points are on z=0 and z=1. If we draw the graph of the restriction, it should be a solid cylinder along z-axis, with height of 1 and radius of 1, so the boundary point could also be on the side face, not just the top and the bottom. $\endgroup$ – user133140 Mar 23 '14 at 3:36
  • $\begingroup$ The boundary points of the level set where $x^2 + y^2 = 1$ and $0 \leq z \leq 1$ are $z = 0$ and $z = 1$. Think of this as a hollowed out cylinder. If you're not at $z = 0$ or $z = 1$ how can you just walk off the edge of the surface? $\endgroup$ – Chris K Mar 23 '14 at 4:00
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I find the max f = 2/(3^(3/2)), and min f = -2/(3^(3/2)). To see this true let x = rcost, and y = rsint, and consider 0 <= r <= 1. If you look at g(x,y) = yx^2 = r^3(cosx^2*sinx) = r^3(1 - sinx^2)sinx = r^3(sinx - sinx^3). Let h(t) = t - t^3 on [-1, 1] then h'(t) = 1 - 3t^2 = 0 <==> t = +/- 3^(-1/2), and h''(t) = -6t. From this h(t) attains max when t = 3^(-1/2) and attains min when t = -3^(-1/2) and the max h = 2/(3^(3/2)) and min h = -2/(3^(3/2)). So if you let r --> 1 and z --> 1 then max f = max h = 2/(3^(3/2)) and min f = min h = -2/(3^(3/2)).

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  • $\begingroup$ Can you please edit your post to use proper formatting? It's very hard to follow your post without it. $\endgroup$ – ml0105 Mar 23 '14 at 2:49
  • $\begingroup$ @ml0105: let me try $\endgroup$ – DeepSea Mar 23 '14 at 2:50
  • $\begingroup$ This is a easier way, I got it, thanks. $\endgroup$ – user133140 Mar 23 '14 at 2:58
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For a maximization problem, it's pretty safe to assume $z = 1$; as if $z \in [0, 1)$, $x^{2}yz < x^{2}y$. Substitution is a good approach as well. It allows us to get rid of the $x^{2}$ term. So $f = (1-y^{2})y$. This should be a more feasible problem to maximize.

The minimization problem is equivalent to maximizing $-f = -x^{2}yz$. I would again assume $z = 1$. I'll let you give this one a try.

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  • $\begingroup$ Wow, I never considered this way before, I'll try again,thanks. $\endgroup$ – user133140 Mar 23 '14 at 2:55

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