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Problem:

Maximise the volume $V$ of a cuboid shaped box with closed top, fixed surface area $S$, and side lengths $x, y,$ and $z$

What I've got so far:

$V=xyz$, $S=2(xy+yz+zx)$, $\nabla V = \lambda \nabla P$

and so

$$ \left\{ \begin{array}{c} \partial V / \partial x =\lambda\ \partial g / \partial x \\ \partial V / \partial y =\lambda\ \partial g / \partial y \\ \partial V / \partial z =\lambda\ \partial g / \partial z \\ S =2(xy+yz+zx) \end{array} \right. $$

i.e. $$ \left\{ \begin{array}{c} yz =2\lambda\ (y+z) \\ xz =2\lambda\ (x+z)\\ xy =2\lambda\ (x+y) \\ S =2(xy+yz+zx) \end{array} \right. $$

How do I solve this set of equations? Elimination and matrix methods didn't work. And how could I extrapolate this problem to

  1. a box with no lid?
  2. A closed box of variable shape (i.e. prove the sphere has the lowest surface area to volume ratio)?
  3. A box in higher dimensions ?
Pie
  • 143

2 Answers2

1

$$ \left\{ \begin{array}{c} yz =2\lambda\ (y+z) \ \quad \color{blue}{(I)} \\ xz =2\lambda\ (x+z) \quad \color{blue}{(II)} \\ xy =2\lambda\ (x+y) \quad \color{blue}{(III)} \\ S =2(xy+yz+zx) \quad \color{blue}{(IV)} \end{array} \right. $$

Hint: Divide $(I)$ by $(II)$. Then multiply both sides by the denominators (in brackets). Then you can multiply out the brackets. You will see the relation between x and y, if $z \neq 0$. I hope you can go on from here.

callculus42
  • 30,550
  • Thanks! I managed to solve the equations, getting x=y=z and so $S=6x^2$ and $V=x^3$ (as expected). For the additional questions, I think I've got 1) (answer posted below) but i'm still not sure how I'd generalise the shape for 2)? And for 3), would it work to just add more variables? – Pie Aug 06 '14 at 07:30
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  1. A box with no lid?

The equations become $$ \left\{ \begin{array}{c} yz =\lambda\ (y+2z) \\ xz =\lambda\ (x+2z)\\ xy =2\lambda\ (x+y) \\ S = xy+2yz+2zx \end{array} \right. $$

giving for the maximum: $x=y=2z$ and so $S=3x^2, V=x^3/2$

Pie
  • 143