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We are asked to find the maximum of $$f(x,y,z) = x+2y+3z$$ in the region in $\mathbb R^3$ where $$g(x,y,z) = x^2 +y^2 +z^2\leq w$$ as a function of w.

I've found the critical point (1,2,3) and their value for f. I've then taken the g term and singled out the x value, replacing x in f with the new value for x. I subsequently took the partial derivative of both y and z.

The answer I get is that y=z.

I put this into g and find a critical point, where the value of f is: $$f(w-5y^2, h^{-1/2}, h^{-1/2}) = w-5y^2 + 5h^{-1/2}$$ where $$h=({w-2y^2-3z^2})^{-1/2}$$

How is possible to determine the maximum from that?

amWhy
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Edward
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  • How did you find the critical point $(1,2,3)$? – Nick D. Nov 20 '14 at 21:33
  • Taking the partial derivatives of f? – Edward Nov 22 '14 at 18:26
  • Right, and then you set $\nabla f=0$, i.e. $(1,2,3)=(0,0,0)$, so it looks like there are no critical points $anywhere$, i.e. the maximum will occur on the boundary of the region in question, which is all points $(x,y,z)$ such that $x^2+y^2+z^2=w$. Now the problem basically reduces to finding the maximum of $f$ with a given constraint. – Nick D. Nov 22 '14 at 18:31
  • Okay thank you, so there are no critical points on the interior and only on the boundary. I still don't understand how to pinpoint the maximum in terms of w. Maybe I've already done so? – Edward Nov 22 '14 at 18:37
  • Are you familiar with Lagrange multipliers? I'll put it in an answer. – Nick D. Nov 22 '14 at 18:44
  • Yes and thanks again. I thought that you don't require Lagrange multipliers for this one - I was trying to solve it from the point of view of finding absolute maxima (w/o Lagrange). – Edward Nov 22 '14 at 18:46
  • You're very welcome. See below! – Nick D. Nov 22 '14 at 19:01

1 Answers1

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We have a function $f(x,y,z)=x+2y+3z$ as well as a constraint $g(x,y,z)=x^2+y^2+z^2=w$.

Then we set $\nabla f=\lambda\nabla g$: $$1=2\lambda x$$ $$2=2\lambda y$$ $$3=2\lambda z$$

We get: $$x=\frac{1}{2\lambda}$$ $$y=\frac{1}{\lambda}$$ $$z=\frac{3}{2\lambda}$$

Substituting, we get: $$(\frac{1}{2\lambda})^2+(\frac{1}{\lambda})^2+(\frac{3}{2\lambda})^2=w$$ which gives $$\lambda=\pm\sqrt{\frac{14}{4w}}$$

So we have either $$(x,y,z)=(\sqrt{\frac{w}{14}},2\sqrt{\frac{w}{14}},3\sqrt{\frac{w}{14}})$$ or $$(x,y,z)=(-\sqrt{\frac{w}{14}},-2\sqrt{\frac{w}{14}},-3\sqrt{\frac{w}{14}})$$

It's clear that $f$ evaluated at the first of these two points is our maximum.

Nick D.
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  • Is it meant to be 14 instead of 13? – Edward Nov 22 '14 at 19:13
  • Haha, yes, adding 1-digit numbers is a problem, I guess. – Nick D. Nov 22 '14 at 19:13
  • Hi @NickD. . I'm trying to prove an identity using Lagrange multipliers and standard vector equations. I'd be very grateful if you even had a hint to point me in the right direction - http://math.stackexchange.com/questions/1034055/lagrange-identity – Edward Nov 23 '14 at 18:50