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July 27th, 2017, 07:22 AM   #1
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Minimize surface area for tank

Hello forum. I am trying to make a rectangular tank that has a volume of 500 cubic feet, but minimizes surface area. The rectangular tank does not have a lid.

$V = xyz = 500$ $\Rightarrow z = \frac{500}{xy}$

Surface area can be described as: $S = 2yz + 2xz + xy$ This accounts for the walls and base, but not the lid.

Plugging in the constraint formula.

$S = 2y(\frac{500}{xy}) + 2x(\frac{500}{xy}) + xy$ $\Rightarrow \frac{1000}{x} + \frac{1000}{y} + xy$

This is now a function with respect to two variables.

$S(x,y) = \frac{1000}{x} + \frac{1000}{y} + xy$

$\frac{\partial S}{\partial x}$ = $-\frac{1000}{x^2} + y$

$\frac{\partial S}{\partial y}$ = $-\frac{1000}{y^2} + x$

Setting these partials equal to $0$.
$-\frac{1000}{x^2} + y = 0$ $\Rightarrow y = \frac{1000}{x^2}$
and...
$-\frac{1000}{y^2} + x = 0$ $\Rightarrow x = \frac{1000}{y^2}$

At this point I hit a wall. I struggle in actually finding the critical point here. It seems that x and y will have the same values though. Perhaps my tank should be a cube. Could anyone help me out with this problem? Am I on the right track?
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July 27th, 2017, 07:53 AM   #2
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$x = 1000/(1000/x^2)^2 = x^4/1000$, so $x = 10$.
Similarly, $y = 10$, and so $z = 5$.
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July 27th, 2017, 08:16 AM   #3
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Quote:
Originally Posted by skipjack View Post
$x = 1000/(1000/x^2)^2 = x^4/1000$, so $x = 10$.
Similarly, $y = 10$, and so $z = 5$.
Thanks a ton for clearing this up, sir!
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