Question

There's an optimization problem for my AP Calculus class that I am stuck on:

You just adopted an iguana and are trying to build a pen with a rectangular bottom and 4 sides with the largest possible volume (essentially an open-top box). Using only 75 feet of building materials, what dimensions will produce a pen with maximum volume?

A diagram is attached.

Thank you!!

Answer 1

x = 5 ft

y = 5 ft

h = 2.5 ft

Explanation:

The surface area of the base of the box will be at its maximum when the difference between the length and width at its minimum.

Therefore, let y = x.

Therefore, the equation for the surface area of box is:

`\implies A=xy+2xh+2yh`

Substituting y = x:

`\implies A=x^2+2xh+2xh`

`\implies A=x^2+4xh`

Given the surface area of the box is 75 square feet:

`\implies 75=x^2+4xh`

Rearrange the equation to create an expression for h in terms of x:

`\implies h=(75-x^2)/(4x)`

The equation for the volume of the box is:

`\implies V=xyh`

Substitute y = x and the expression for h to create an equation for volume in terms of x:

`\implies V= (x^2(75-x^2))/(4x)`

`\implies V= (x(75-x^2))/(4)`

`\implies V= (75)/(4)x-(1)/(4)x^3`

To find the value of x that maximizes the volume, differentiate V with respect to x and find the value(s) of x that makes dV/dx = 0.

`\implies \frac{\text{d}V}{\text{d}x}=(75)/(4)-(3)/(4)x^2`

Set it to zero and solve for x:

`\implies (75)/(4)-(3)/(4)x^2=0`

`\implies (3)/(4)x^2=(75)/(4)`

`\implies 3x^2=75`

`\implies x^2=25`

`\implies x=5`

Check to see if this value of x gives a minimum for V by inputting it into the second derivative:

`\implies \frac{\text{d}^2V}{\text{d}x^2}=-(3)/(2)x`

`x=5\implies \frac{\text{d}^2V}{\text{d}x^2}=-(3)/(2)(5)=-(15)/(2) < 0 \implies \sf maximum`

Finally, substitute the found value of x into the expression for h to find the height of the box:

`x=5 \implies h=(75-(5)^2)/(4(5))=(5)/(2)=2.5`

Therefore, the dimensions that will produce a pen with maximum volume are:

• x = 5 ft
• y = 5 ft
• h = 2.5 ft