Maximize rabbit pen size (no calc)

Question

Question:

To make an enclosure for your pet rabbits, you want to fence a rectangular pen against your house. Only 3 sides will need to be fenced. You have 170 feet of fencing material, and want to maximize the pen area.

a. Write a quadratic function (in standard form) to represent the area of the rectangular enclosure, where x is the width.

b. What dimensions will maximize the area of the enclosure?

c. What is the maximum area of the enclosure?

I believe that this type of problem can be solved with calculus, but I am in an algebra class, so please avoid the use of calculus.

My Attempt:

I'm pretty lost, when it comes to this problem. My best guess is that the Area$=(170/3)^2=28900/9$, since a square has the largest area among quadrilaterals with equal perimeters. I really have no clue how to solve this, though.

Thanks in advance

Answer 1

Start with putting the given information into equation form:

Total length of fence: L = 170.

Area enclosed: S = x*y,

where x is the length transverse to the house and y is the length parallel to the house.

In terms of the sides, the total length can be written as:

L = y + 2x = 170,

because there is one side parallel to the house and two sides transverse to the house.

We want to relate this to the area x*y so we can conveniently multiply everything in the equation for L by x to get:

$x y + 2 x^2 = 170 x$

$S + 2 x^2 = 170 x$

$S = 170 x - 2x^2 $

Put this into standard quadratic form by equating everything to zero

$-2 x^2 +170 x - S = 0$

This answers question (a).

Now back to the area formula we derived above:

$S = 170 x - 2x^2 $

If you plot this equation for the area you will see the maximum or peek value is between the two roots where the curve crosses the x axis, due the symmetry of the curve.

The quadratic formula for solving for the roots is:

$\frac{-B \pm \sqrt{B^2 - 4 A C}}{2 A}$

where A = -2, B = 170 and C = S (Area) from our standard form quadratic.

If we rewrite the quadratic equation as:

$\frac{-B}{2 A} \pm \frac{\sqrt{B^2 - 4 A C}}{2 A}$,

it is easy to see that the midpoint between the two roots is simply: $$\frac{-B}{2A}$$

This is the x value when the area is at a maximum. We can find the area by using this value of x in the $S = 170 x - 2x^2 $ equation we obtained earlier. This answers question (b).

The y value can found by substituting x into the $y + 2x = 170$ equation we found earlier. This answers question (c)

Answer 2

Geometric solution:

Cutting the rectangle in half (splitting perpendicular to the open side), leaves two identical rectangles, both with a height of $y$ and a base of $x/2$ and the perimeter restriction is equivalent to the sum of the width and length of each piece being $85$. As you know, a square maximizes area when the sum of the width and length (half the perimeter) is fixed, so the best shape for them is $85/2$ by $85/2$, so the best shape is 85 by 42.5.