The wording confuses me

Question

60 meters of fencing is available to enclose a pig pen along the side of a barn. What dimensions of the pen with maximize the area of the pig pen? Only three sides of the pen need to be fenced in as the barn will act as a barrier for one side.

Let x represent the width an let 60-2x represent the side opposite the barn.

Answer 1

So this is just an optimization problem. The first thing you want to do is to try to visualize the pen. Since the pen is fenced against the barn, the 60 meters of fencing is used for only three sides of the pig pen. Once you draw out the pen, the two opposite sides can be labelled as x and the other labelled as 60-2x.

Now, since we want to maximize the area, we are basically trying to optimize this expression:

A(x) = x(60-2x)

A(x) = 60x - 20x²

We then take the derivative of this using the product rule, and we get:

A'(x) = 60 - 4x

Next, we must find the critical points of the derivative. This would show us the maximum or minimum points of the original equation, A(x). If this doesn't make sense, think of a graph and its derivative graph. The derivative graph always shows the original graph's minimum or maximum points at the x-axis or if it's undefined. Well, that is exactly what we are going to do. We need to find the critical points by solving for A'(x) = 0 or A'(x) = undefined.

A'(x) = 0 A'(x) ≠ undefined

60 - 4x = 0

x = 15

To make sure that this is the maximum, we must use a number line. You basically put 15 in the middle of the number line and pick a number less than 15 and see if it's a positive number when you plug it back into the original equation. Then, you pick a number greater than 15 and see if it's a negative number. If it's true, then 15 is the maximum dimension for x. Thus the maximized area is 60(15)-2(15)² = 450.

Answer 2

Using calculus, you may also verify that the 2nd derivative is negative for a maximum.

Since A'(x) = 60 - 4x, A"(x) = -4, which is indeed < 0, indicating a maximum.

According to the problem header, this is an Algebra 2 problem. But Jahnavi's solution and mine use calculus. You seem to be OK with that. However, let us know if you need a non-calculus solution, which I would suggest involves parabola properties, and can get it later for you. But I need some sleep now!