Максимізація прямокутної площі з огорожею

Problem

A farmer builds a rectangular pen against a barn wall with $300$ feet of fencing, using three fenced sides with bottom length $x$ and two equal ends of length $y$; what dimensions maximize the area?

Step 1: Write the fencing constraint

Since only three sides need fencing, the total fencing gives

$$x + 2y = 300.$$

Solving for $y$ in terms of $x$ gives

$$y = \frac{300 - x}{2}.$$

Step 2: Express area as a function of one variable

The area of the rectangle is

$$A = xy.$$

Substitute $y = \dfrac{300-x}{2}$ to get

$$A(x) = x \cdot \frac{300-x}{2} = 150x - \frac{x^2}{2}.$$

This is a downward-opening parabola in $x$.

Step 3: Differentiate and find the peak

Differentiate the area function:

$$A'(x) = 150 - x.$$

Set the derivative equal to $0$:

$$150 - x = 0.$$

So

$$x = 150.$$

Then

$$y = \frac{300 - 150}{2} = 75.$$

Step 4: Compute the maximum area

The maximizing dimensions are $x = 150$ feet and $y = 75$ feet, so the maximum area is

$$150 \cdot 75 = 11{,}250$$

square feet.

Answer

The maximum area is $11{,}250$ square feet, achieved when the pen measures $150$ feet by $75$ feet.