Hung Lam plans to enclose a rectangular area for his chickens in his backyard using the sides of his apartment building for $2$ of the sides, as seen in the figure. If Hung Lam has exactly $40$ feet of chicken wire, what is the largest area, in square feet, he can enclose for his chickens?

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Q. Hung Lam plans to enclose a rectangular area for his chickens in his backyard using the sides of his apartment building for

2

of the sides, as seen in the figure. If Hung Lam has exactly

40

feet of chicken wire, what is the largest area, in square feet, he can enclose for his chickens?

Understand the problem: Understand the problem. $\newline$ Hung Lam is using the sides of his apartment building as two sides of the rectangular enclosure, which means he only needs to use the chicken wire for the remaining two sides. Since he has $40$ feet of chicken wire, this will be the perimeter of the two sides that he needs to enclose with the wire.
Determine side lengths: Determine the lengths of the sides that need to be enclosed with chicken wire. Let's denote the length of the side parallel to the apartment building as $L$ and the length of the side perpendicular to the apartment building as $W$ . Since there are two sides of the rectangle that need chicken wire, the total length of chicken wire used will be $2W + L = 40$ feet.
Express area in terms of one variable: Express the area of the rectangle in terms of one variable. $\newline$ The area $A$ of the rectangle can be expressed as $A = L \times W$ . However, we need to express $L$ in terms of $W$ using the perimeter equation from Step $2$ . Rearranging the equation gives us $L = 40 - 2W$ .
Substitute into area formula: Substitute the expression for $L$ into the area formula. $\newline$ Now we can express the area solely in terms of $W$ : $A = (40 - 2W) \times W$ .
Find value of W: Find the value of $W$ that maximizes the area. $\newline$ To maximize the area, we need to find the value of $W$ that makes the derivative of the area function equal to zero. The area function is a quadratic function, and its maximum value occurs at the vertex of the parabola. For a quadratic function in the form of $A = -kW^2 + bW$ , the vertex occurs at $W = \frac{b}{2k}$ . In our case, $A = -2W^2 + 40W$ , so $k = 2$ and $b = 40$ . Therefore, the value of $W$ that maximizes the area is $W = \frac{40}{2 \times 2} = \frac{40}{4} = 10$ feet.
Calculate maximum area: Calculate the maximum area using the value of $W$ . Substitute $W = 10$ feet back into the area formula: $A = (40 - 2 \times 10) \times 10 = (40 - 20) \times 10 = 20 \times 10 = 200$ square feet.