Question

a. a rectangular pen is built with one side against a barn. if 100 m of fencing are used for the other three sides of the pen, what dimensions maximize the area of the pen?
b. a rancher plans to make four identical and adjacent rectangular pens against a barn, each with an area of 25 m². what are the dimensions of each pen that minimize the amount of fence that must be used?
a. to maximize the area of the pen, the sides perpendicular to the barn should be □ m long and the side parallel to the barn should be □ m long.
(type exact answers, using radicals as needed.)

Explanation:

Step1: Let the side perpendicular to the barn be \(x\) and the side parallel to the barn be \(y\)

For part a, the total length of the fence is \(2x + y=100\), so \(y = 100 - 2x\). The area \(A=xy=x(100 - 2x)=100x-2x^{2}\)

Step2: Find the derivative of the area function

Differentiate \(A(x)\) with respect to \(x\). \(A^\prime(x)=\frac{d}{dx}(100x - 2x^{2})=100-4x\)

Step3: Set the derivative equal to zero to find critical points

Set \(A^\prime(x)=0\), then \(100 - 4x=0\). Solving for \(x\):
\(4x=100\), \(x = 25\)

Step4: Find the second - derivative

\(A^{\prime\prime}(x)=\frac{d}{dx}(100 - 4x)=-4<0\). Since \(A^{\prime\prime}(x)<0\) when \(x = 25\), the function \(A(x)\) has a maximum at \(x = 25\)
Substitute \(x = 25\) into \(y=100 - 2x\), \(y=100-2\times25 = 50\)

For part b, let the side perpendicular to the barn be \(x\) and the side parallel to the barn be \(y\). The area of each pen is \(A=x\times\frac{y}{4}=25\), so \(y=\frac{100}{x}\). The total length of the fence \(L = 5x+y\). Substitute \(y=\frac{100}{x}\) into \(L\), we get \(L(x)=5x+\frac{100}{x}\)

Step5: Find the derivative of the fence - length function

Differentiate \(L(x)\) with respect to \(x\). \(L^\prime(x)=5-\frac{100}{x^{2}}\)

Step6: Set the derivative equal to zero to find critical points

Set \(L^\prime(x)=0\), then \(5-\frac{100}{x^{2}}=0\). \(5x^{2}=100\), \(x^{2} = 20\), \(x = 2\sqrt{5}\)
Substitute \(x = 2\sqrt{5}\) into \(y=\frac{100}{x}\), \(y = 10\sqrt{5}\)

Answer:

a. The sides perpendicular to the barn should be \(25\) m long and the side parallel to the barn should be \(50\) m long.
b. The dimensions of each pen: the side perpendicular to the barn is \(2\sqrt{5}\text{ m}\) and the side parallel to the barn is \(10\sqrt{5}\text{ m}\)