July 27th, 2017, 07:22 AM  #1 
Senior Member Joined: Nov 2015 From: United States of America Posts: 198 Thanks: 25 Math Focus: Calculus and Physics  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? 
July 27th, 2017, 07:53 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,805 Thanks: 2150 
$x = 1000/(1000/x^2)^2 = x^4/1000$, so $x = 10$. Similarly, $y = 10$, and so $z = 5$. 
July 27th, 2017, 08:16 AM  #3 
Senior Member Joined: Nov 2015 From: United States of America Posts: 198 Thanks: 25 Math Focus: Calculus and Physics  

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