Question

A farmer has 300 ft of fencing with which to enclose a rectangular pen next to a barn. The barn itself will be used as one of the sides of the enclosed area.
What is the maximum area that can be enclosed by the fencing?

_______ft^2

Answer 1

This question can be answered in a number of ways depending on your level of mathematical sophistication (using calculus, graphing a parabola, and guess/check). I'll talk through all three in the order above.

They all start the same way: Let's call the length of the rectangle b and the width a. The perimeter of the rectangle (the distance all the way around -- that is, the length of the "fence") is 2a + 2b. Since the barn is used as one of the sides (let's say b) we can subtract b ... we don't need fencing for this side. That makes the perimeter 2a + b. Since we have 300 feet of fencing we set these equal: 2a + b = 300.

We can solve for b above by subtracting 2a from both sides of the equation so we get: b = 300 - 2a.

We are asked to maximize the area. The area of a rectangle is length times width. In this case, (a)(b) and since we just found an expression for b, we can substitute this to obtain: Area = $(a)(300-2a)=300a-2 a^{2}$. This is a parabola when graphed and we are asked to find where it's highest point occurs because this is where the area is greatest.

METHOD 1: Using Calculus
We are looking for the highest point of the equation $Area = [tex](a)(300-2a)=300a-2 a^{2}$. Let's call the function R(a) since it is in terms of a. To find the maximum we take the first derivative, set it equal to zero and solve for a. As this is a parabola (with the leading coefficient being negative) it is concave down and the point we find will be the only maximum.

First we find the derivative of the function.
$R(a)=(a)(300-2a)=300a-2 a^{2} [tex]\\Next set it equal to 0 and solve for a. R^{'}(a)=300-4a 0=300-4a -300=-4a 75 = a$ [/tex]
This means that the maximum value (the largest area) occurs when a = 75.
At a = 75 the area is given by
What is the largest area? Let's us substitute 75 for a in the equation of the area. This gives: $Area=a(300-2a)=(75)(300-(2)(75))=(75)(150)=11250$ square feet.

METHOD 2: Graphing
We know that the area is given by. Area = $(a)(300-2a)=300a-2 a^{2}$. To make this more familiar let's use y for the area and x for the width (a). That means we have the equation: $(a)(300-2a)=300a-2 a^{2}$.

This is a quadratic equation (the highest exponent of x is 2) and so the equation of a parabola (like the letter U or an upside down U). Since the leading coefficient is negative (-2) we know it opens downward (called concave down -- think of a frown). We are looking for the x-coordinate of the highest point called the vertex. We can find this by remembering that the x-coordinate of the vertex is given by $x= \frac{-b}{2a}$ where a is the leading coefficient the coefficient of the term with the x squared (here = -2) and a is the coefficient of the linear term (the one with x...here 300). That makes the x-coordinate of the vertex $x= \frac{-300}{2(-2)}= \frac{-300}{-4}=75$.

The y-coordinate of the vertex is found by substituting 75 for x in the equation for the area. $y=300x-2 x^{2} y=300(75)-2(75)^{2}$
$y=22500- 2(5625) y=22500- 11250 y=11250$

Since we let x be the width of the rectangle. We know that the width is 75. The area of the rectangle we called y. So the maximum area is 11250 square feet which occurs when the width of the rectangle is 75.

You could have gotten to this point by generating a list of points for the parabola and graphing it carefully on graph paper (though it might not have been exact depending on how you scale your axis).

METHOD 3: Guess and Check
The area of the rectangle is given by Area = $(a)(300-2a)=300a-2 a^{2}$. Here you guess values for a, plug them into the formula to get the area and keep doing this until you are comfortable you have found the largest area. Not the best method but if it's a test and you forget everything else, it might get you close to an answer or even get you the exact answer (depending on how well you guess). It might take a while so I'd do this question last if this were a test.