What dimensions will produce a box with the largest volume?

Question

A manufacturer wants to design an open top box with a square base having a surface area of 108 square inches.

Answer 1

So, in this problem, you are trying to maximize the volume of the box. These are known as optimization problems. To solve this problem, you will need two equations. The first equation is the volume formula, the quantity to are wanting to optimize. This is often referred to as the primary equation. The other equation comes from the given information. In this case, the information given is related to the surface area of the box.

Primary Equation: V = s²h Secondary equation: s² + 4sh = 108

We want to differentiate the primary equation and find critical values in hopes of determining the dimensions that will yield a maximum volume. Before we can do that, we need volume defined in terms of a single variable. That is where the secondary equation comes in: Solve it for h and then substitute for h in the volume equation:

h = (108 - s²)/4s then

V = s²[(108 - s²)/4s]

Simplify to make it easier to differentiate:

V = 27s - .25s³

So dV/ds = 27 - .75s²

Next, find the critical values:

27 - .75s² = 0

.75s² = 27

s² = 36

s = 6 or s = -6, we only need to check s = 6 since s = -6 does not make sense in the context of this problem.

Verify that s = 6 would give the maximum volume by checking to see that the derivative is positive for any values smaller than s = 6 and negative for any values larger than s = 6:

V(1) = 27 - .75(1) > 0 and V(8) = 27 - .75(64) < 0

Therefore, s = 6 does give the maximum volume. To find the corresponding height of the box, substitute s = 6 back into the secondary equation and solve for h:

6² + 4(6)h = 108

36 + 24h = 108

h = 3

So, the dimensions that will produce the largest volume are s = 6 in. and h = 3 in.

The maximum volume is V = 6² · 3 = 36 · 3 = 108 in².

Answer 2

Dear King,

first of all sketch the box. Label with x the sides of the squared base and with h the height.

The surface area will be given by Area of the bottom base (x · x) + area of the 4 lateral surfaces (each is a rectangle with an area of x · h).

This gives us:

108 = x² + 4xh.

In this problem we want to optimize the volume, meaning we want to find x so that the volume is maximum.

The volume is given by:

V = Area of the base · height = x²· h.

This function has to be only in terms of x, therefore I use the previous function to solve for h and then substitute into the volume.

4xh = 108 - x²

h = (108/4x) - (x²)/4x = 27/x - (1/4)x

Substituting:

V = x² (27/x - (1/4)x) = 27x - (1/4)x³

To find the maximum value, we need to find now the derivative:

V' = 27 - (3/4)x².

Let's find the critical numbers by setting V' = 0

27 - (3/4)x² = 0, solving for x we find x² = 36, which gives x = 6 (it cannot be –6 because x is a dimension).

Substitute into h = 27/(6) - (1/4)(6) = 3

Therefore the dimensions are 6 by 3.

I hope that this is helpful, and please contact me if you need more help in Calculus. I love teaching Calculus!

Answer 3

Hi King K.,

This is a fairly standard calculus problem. You want to maximize the volume, and you only control .... What dimension?

Turns out, you only NEED to specify one dimension as a variable. Let's call it one side of the square base, and designate it as b (in inches).

Then the bottom has area b^2, and the 4 identical box sides have area (each) of (108-b^2)/4 . Right? That will add up to 108 square inches, yes?

So what is the height of the box? If a box side has area as above, and one edge is length b, then you may calculate the other edge, I think.

You now have all 3 dimensions of the box, all functions of b. So you take the product of those three functions, which is the volume, and differentiate once with respect to b. Why? Because you KNOW that there must be an optimum, and only one!, and that derivative will be zero for that value of b.

Then solve the differential expression = 0 for roots in terms of b.

This neglects, of course, that standard boxes have considerable overlap of flaps on the bottom, and a slight overlap on a vertical edge, or that they might be wrinkled.

........ But how do you get a box with no wrinkles?

[Start] " Wrinkle to(->) box" --> "Wrinkle to(->)box" --> "Wrinkle bto(->)ox" --> " botox" [End]

-- Cheers, -- Mr. d.