Question

A small shelter for delicate plants is to be constructed of thin plastic material. It will have a square end and a rectangular top and back, with an open bottom and front as shown in the figure. The total are of the four plastic sides is to be 1200in^2.
a) Express the volume of the shelter as a function of the depth x.
b) What dimensions will maximize the volume of the shelter?

Answer 1

Part A

The four plastic sides comprises of two equal squares of area $x^{2}$ each and two equal rectangles of area $xy$ each.

Given that the total area of the four plastic sides is to be $1200in^2$, thus:

$2x^2+2xy=1200 \\ \\ \Rightarrow x^2+xy=600 \\ \\ \Rightarrow xy=600-x^2$

The volume of the figure is given by: Volume = Area of base x depth

$V=(xy)x=x(600-x^2)=600x-x^3$



Part B:

For maximum volume, the derivative of V with respect to x will equal 0.

$\frac{dV}{dx} =0 \\ \\ \Rightarrow600-3x^2=0 \\ \\ \Rightarrow3x^2=600 \\ \\ \Rightarrow x^2=200 \\ \\ \Rightarrow x=\pm \sqrt{200} =\pm14.14$

But dimensions has to be positive, thus the value of x which produces maximum volume is x = 14.14

Recall that

$xy=600-x \\ \\ \Rightarrow y= \frac{600}{x} -x \\ \\ = \frac{600}{14.14} -14.14 \\ \\ =42.43-14.14 \\ \\ =28.29$

Therefore, the dimensions that will maximize the volume of the shelter is x = 14.14 and y = 28.29