A cardboard box (this one has no top) will have a square base and needs to enclose 40,000 cm^3. What should the height of the box be, if it is going to use the least cardboard?

Answer 1

The volume of this box is a given constant, but for ease we'll let it be represented by a variable V. Let b denote the length of one of the (equal) sides of the base, and let h denote the height. The box has to enclose volume V, so we know that we have the constraint

V = b^2 x h

so we can express b^2 = V/h.

Now let's write an expression for the amount of cardboard we'll use to construct the box. Let's denote this A for surface Area. If you sketch out the cube, you'll see that we have one base with area b^2 and four sides of area bxh (and no top, as you said).

So, we can express the area as

A = b^2 + 4 b x h

We can substitute in our expression for b^2 to get a function for area in terms of h:

A(h) = (V/h) + 4 b x h

Now we want to minimize A. If there's some minimum to this function, it'll have the property that dA/dh = 0, so let's solve for values of h that would make this happen:

dA/dh = -V/h^2 + 4 b = 0

4b = V/h^2

h = square root (V/4b).

Now we can plug in our numerical value of V to find our answer for the height in terms of the length of the side of the base.

Something for you to think about: how can you check that this solution dA/dh = 0 is a minimum, and not a maximum?