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Show that a box (rectangular parallelopiped) of maximum volume V with prescribed surface area is a cube. Let $$V=xyz$$ $$S=2xy + 2yz + 2zx$$ $S$ is constant.

Using Lagrange method, I am stuck at $V_x$$_x$=$0$=$V_y$$_y$=$V_z$$_z$ at the (only) critical point. How to approach this.

T.Pal
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2 Answers2

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HINT: Lagrange multiplier $ \lambda $

$$ \frac{V_x}{S_x} =\frac{V_y}{S_y}= \frac{V_z}{S_z} = -\lambda $$

Narasimham
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  • Thanks. But I have done similarly. I have got (a/6)^1/2 = x = y =z where 'a' is the constant surface area. This critical point gives 0=fxx=fyy=fzz how to show that this sole critical point is maxima? – T.Pal Mar 26 '16 at 19:30
  • Does Lagrange method always result in maxima or minima? For e.g. if there are 3 critical points, they necessarily will be either minima or maxima? @Narasimham – T.Pal Mar 27 '16 at 13:13
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Of course you can do this with Lagrange multipliers. But the AM-GM-inequalitiy suffices: $$V^{2/3}=(xy\cdot yz\cdot zx)^{1/3}\leq{1\over3}(xy+yz+zx)={1\over6}S$$ with equality sign iff $x=y=z$.

Concerning Lagrange multipliers: Look at second derivatives only "in extremis". For a standard problem use qualitative reasoning to prove that the point produced by the method is actually the global extremum you are looking for.

Christian Blatter
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  • Oh this is quite innovative. Thanks. :) But I am not able to get through with Lagrange multiplier. The double derivative wrt x,y,x at critical point comes out to be zero. Not sure how to go and show that the only critical point is maxima. – T.Pal Mar 26 '16 at 20:10
  • Fair enough! So I think I would just calculate the critical point and use AM-GM inequaity and since in my (only) critical point x=y=z, the AM-GM thing becomes an equality and that suffices for it being a cube! Thanks a ton! :) – T.Pal Mar 26 '16 at 20:24
  • Does Lagrange method always result in maxima or minima? For e.g. if there are 3 critical points, they necessarily will be either minima or maxima? @Christian Blatter. Can there not be any saddle point? – T.Pal Mar 27 '16 at 13:13
  • @christian Blatter ;what if double derivative comes out to be 0 at critical points and question isn't of standard form . then how will we come to know weather point yield maximum or minimum – Naruto_007 Aug 05 '17 at 11:12
  • @Naruto_007: Lagrange's method produces conditionally stationary points, period. There is an analogue of the Hessian also in this case, but it involves second derivatives of the constraint as well. This is usually not taught in calculus, and I strongly advise not going into it. – If the (conditionally) stationary point is degenerate a complete diagnosis of its character is a very difficult task even in the nonconditional case. – Christian Blatter Aug 05 '17 at 11:49