Word Problems – 4

Continuing this series on word problems, let’s look at one that many year 11 or 12 students have seen if they have covered calculus. The first part of the question though, does not need calculus:

A 6 by 8 cm rectangular piece of metal has a square cut out of each corner:

The metal is then folded along the dashed lines to form a box of height x.

a) What is the volume of the box in terms of x?

b) What is the maximum volume the box can have and at what value of x does the maximum occur?

Let’s first redraw the rectangle, labelling what we know. If the height of the box is to be x, then that is the size of the cutout square. If an x is subtracted from each end of each side, then the length of each of the dotted lines (the base of the future box) is 6 – 2x and 8 – 2x:

After the metal is folded, we have a box like:

So to answer part a), the volume is height × length × width. So

V = x(8 – 2x)(6 – 2x)

For part b), we need a little calculus. But before we do that, just to get a mental image of what is going on (this is not needed to solve the problem though), let’s plot V as a function of x:

Notice that the volume is 0 at x = 0 and x = 3. This makes physical sense because when x = 0, there is no square cut out and the box would just be a flat sheet. If x = 3, there would be nothing left of the original 6 cm side and the two remaining flaps would just fold up against each other giving no volume. So the physical restriction on x here is 0 ≤ x ≤ 3.

The maximum volume then occurs at the peak of the curve between 0 and 3. To find what and where this maximum is, requires calculus, specifically finding a local maximum (a stationary point).

To find this point, let’s first expand the expression for V:

V = x(8 – 2x)(6 – 2x) = 4x3 – 28x2 + 48x

For the uninitiated, derivatives of a function give the gradients of lines tangent (that is just touching at one point) to it. At the maximum, the tangent line is horizontal which has a gradient of 0. So to find this, we find the derivative of our volume function and set that equal to zero and solve for x.

For those who have had calculus and know how to find derivatives of a polynomial function (like we have here):

V′ = 12x2 – 56x + 48 = 4(3x2 – 14x + 12)

So now we find the values of x where the derivative is 0:

V′ = 4(3x2 – 14x + 12) = 0 ⟹ x = 1.1315, 3.5352

To get that answer, the quadratic formula can be used. Or, if lazy like me, an equation solver on the internet.

We want the value between 0 and 3. So a square of sides 1.1315 cm will maximise the volume. The other value of x is where the minimum point is as seen in the plot. We find the maximum volume by putting x = 1.1315 in the original V function:

V(1.1315) = 4(1.1315)3 – 28(1.1315)2 + 48(1.1315) = 24.258 cm3

As before, drawing pictures gets you started.