Part A
The four plastic sides comprises of two equal squares of area [tex] x^{2} [/tex] each and two equal rectangles of area [tex]xy[/tex] each.
Given that the total area of the four plastic sides is to be [tex]1200in^2[/tex], thus:
[tex]2x^2+2xy=1200 \\ \\ \Rightarrow x^2+xy=600 \\ \\ \Rightarrow xy=600-x^2[/tex]
The volume of the figure is given by: Volume = Area of base x depth
[tex]V=(xy)x=x(600-x^2)=600x-x^3[/tex]
Part B:
For maximum volume, the derivative of V with respect to x will equal 0.
[tex] \frac{dV}{dx} =0 \\ \\ \Rightarrow600-3x^2=0 \\ \\ \Rightarrow3x^2=600 \\ \\ \Rightarrow x^2=200 \\ \\ \Rightarrow x=\pm \sqrt{200} =\pm14.14[/tex]
But dimensions has to be positive, thus the value of x which produces maximum volume is x = 14.14
Recall that
[tex]xy=600-x \\ \\ \Rightarrow y= \frac{600}{x} -x \\ \\ = \frac{600}{14.14} -14.14 \\ \\ =42.43-14.14 \\ \\ =28.29[/tex]
Therefore, the dimensions that will maximize the volume of the shelter is x = 14.14 and y = 28.29
