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I appreciate you taking the time to assist, I will not waste time, jump right into this question; very tricky! I can solve it in 1 minute with basic multiplication/division, but I'm trying to figure it out using Derivatives.

You are to design an open top rectangular stainless steel vat. It is to have a square base and a volume of 64 feet cubed. What are the dimensions for the base and height (the same, since the bottom of a square) that will make the tank weigh as little as possible.

In calculus you want to find a function and takes it's first derivative and find a minimum. You want the derivative to be decreasing on an interval and then begin increasing. So picture a graph going down left to right and then it suddenly starts to increase. The point at where it changed from decreasing to increasing is what is called a local minimum.

Now where do you get the function?

The volume of a rectangular solid is lwh but the base is square so let the base have length and width x. Call the third dimension the height h. Now you have one equation of V = x^2h but you know the volume is 64 cubic feet so you really have 64 = x^2h. Solve for h getting h=64/x^2.

You need another relationship to make a function f that you can take the first derivative of. Since you are trying to minimize the tanks weight, use it's surface area to minimize it.

You will have 4 sides with an area of xh and the base is x^2. Since there is no top you have all the pieces to write the total surface area.

Let the surface area be f(x).

f(x)=4xh + x^2 and substitute for h, 64/x^2.

Thus f(x) = 4x(64/x^2)+x^2 = 256/x+x^2 =>

f(x) = 256x^-1 + x^2

f'(x) = -256x^-2 + 2x = -256/x^2 + 2x

Set f'(x) = 0 and factor to find any critical points (where the function is 0 or undefined).

-256/x^2 + 2x = 0

+256/x^2 +256/x^2

2x =256/x^2 multiply both sides by x^2

(x^2)2x = (x^2)(256/x^2)

2x^3 = 256. divide both sides by 2

x^3 = 128

Take the cube root of each side getting

x = cube root of 128 = 4*square root of 2

(I am using V for the square root)

x = 4V2, this is your only critical point,you need to check it it is a minimum.

Break up the number line with x = 4V2

<--------|--------->

4V2

Choose a number (test point) to the left of the critical point and a number to the right and plug them into f(x) to see if it is negative or positive. You will have a local minimum if the value on the left is negative and the value on the right is positive.

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