Question

Applied Optimization
I just need to know how to get started.
A Rectangular tank that is 6912 ft^3 with a square bbase and open top is to be constructed of sheet ssteel of a given thickness. FIND THE DIMENSIONS OF THE TANK WITH THE MINIMUM WEIGHT.

Answer 1

The tank with minimal weight is the same as the tank with minimal surface area. At least, that's the assumption, which I'm basing off of this fact: The smallest surface-area-tank will use the least material, so it'll weigh the least.

The surface area of a rectangular tank (prism) is given by \(A=\sum(\text{areas of faces}).\) The base of the tank is square, so the length and width are the same. Let \(x\) denote the length of the tank and let \(y\) denote the height of the tank. Thus, \(A=2x^2+4xy.\)

This particular tank has an open top, so you don't consider the area of the face that would normally be there. The area of this face is \(x^2,\) the same as the area of the base. Subtract this from the total area, and you have the following formula:
\[\large A=x^2+4xy\]

The volume of the tank is given by \(V=x^2y,\) which is just the product of length, width and height. You're given the volume - you know it to be \(6912 \text{ ft}^3.\) This gives you the equation
\[\large V=6912=x^2y\]
Now, you want to minimize surface area, which you have in terms of x and y. You'll be using the second equation to find an expression for either x or y (either way works), which you'll substitute into the surface area equation. Then, you minimize the surface area with an application of the first derivative test.