A cuboid has a square base. The height of the cuboid is twice the length of the side of the base. The S.A of the cuboid is increasing at a rate of 10cm^2 s^-1. Find the rate of increase of V, H=12cm.

Question

I'm confused whether the answer is 18 or 36.

Answer 1

Let the height of the cuboid be "h". We know from the problem statement the height is twice the length of the side of the square base. So we basically have a box with dimensions h/2 x h/2 x h (putting everything in terms of one variable "h").

First, we need to come up with an equation for the surface area of the box. The six sided cube will have two square faces at the top and bottom and four rectangular faces on the sides. The equation is written as follows:

SA = (h/2)*(h/2)*(2 sides) + (h/2)*(h)*(4 sides)

SA = (h²/2) + (2h²) = (h²/2) + (4h²/2)

SA = 5h²/2

The other equation we will need to determine is for the volume of the box. This is straightforward (length x width x height):

V = (h/2)*(h/2)*(h)

V = h³/4

Next, let's go back to the surface area equation and calculate the derivative using implicit differentiation with respect to time, t.

dSA/dt = 5h*(dh/dt)

We know the surface area is increasing at a rate of 10 cm²/s. This is "dSA/dt". We also know the height is 12 cm at this particular time. What we do not know is "dh/dt". But we can solve for it.

(10 cm²/s) = 5*(12 cm)*(dh/dt)

(10 cm²/s) = (60 cm)*(dh/dt)

dh/dt = 10/60 = 1/6 cm/s

Ultimately, we are trying to find the rate of increase of V (which is dV/dt). So we will need to implicitly differentiate our volume equation as well.

dV/dt = (3h²/4)*(dh/dt)

We now know everything in the above equation except dV/dt:

dV/dt = [3*(12 cm)²]/4 * (1/6 cm/s)

dV/dt = [3*(144 cm²)]/4 * (1/6 cm/s)

dV/dt = (108 cm²) * (1/6 cm/s)

dV/dt = 18 cm³/s

Notice how the units all work out and we are left with a volumetric rate (cm³/s).