Calculus Optimization

Question

Consider a rectangular box with a volume of 32 ft^3 that has five sides: a bottom and four vertical sides but no top. Let W be the box’s width, L be its length, and H be its height; to be realistic, we assume that W, L, and H are all positive.

Our goal is to find the dimensions of this box that will minimize the surface area. We will find these dimensions through a series of steps.

1)Let A be the surface area of our 32 ft^3 box. Derive a formula for A in terms of only W and L (but not H).

2)First assume that w is fixed (i.e., w is a constant), and view A as a function of L only. Determine the minimum possible surface area of our box.

Note: Since we don't know the value of the fixed-width w, your answer will depend upon w.

3)Since your answer to part (b) depends upon w, it is a function of w; let's denote that answer as A_min(w). (Thus A_min(w) is the minimum possible surface area of our 32ft^3 box when the width is assumed to be w.) Now by viewing w as a variable (i.e., we no longer view w as a constant) and A_min(w) as a function of w, determine the minimum possible value of A_min(w) over all values of w > 0.

4)What are the dimensions of our 32 ft^3 box that will give us the smallest possible surface area?

Answer 1

1. V = WLH, so 32 = WLH

dividing both sides by WL we then get, 

H = 32/WL

A = 2WH + 2LH + WL

Then we substitute H = 32/WL into the equation,

A = 64/L + 64/W + WL

2. If we assume W is a constant we can make it equal to 4, so W = 4

A(L) = 64/L + 16 + 4L

Then to minimize the A(L) we need to find the derivative of A(L) and set it equal to 0

A’(L) = -64/L² + 4 = 0

solving for L we get L = 4, this means that L and W are equal. So L = W = 4

H = 32/16 = 2

3. Now we substitute L = 4 into our Area equation,

A(W) = 64/W + 64/W + WL

A(W) = 128/W + W²

4.

A(4) = 128/4 + 4² = 48 feet squared is the minimum surface area of the box.