a. a rectangular pen is built with one side against a barn. if 200 m of fencing are used for the other three…

a. a rectangular pen is built with one side against a barn. if 200 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 50 m long and the side parallel to the barn should be 100 m long. (type exact answers, using radicals as needed.) b. to minimize the amount of fence that must be used, each of the sides perpendicular to the barn should be m long and each of the sides parallel to the barn should be m long. (type exact answers, using radicals as needed.)

a. a rectangular pen is built with one side against a barn. if 200 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 50 m long and the side parallel to the barn should be 100 m long. (type exact answers, using radicals as needed.) b. to minimize the amount of fence that must be used, each of the sides perpendicular to the barn should be m long and each of the sides parallel to the barn should be m long. (type exact answers, using radicals as needed.)

Answer

Explanation:

Step1: Define variables for part b

Let the side perpendicular to the barn be $x$ meters and the side parallel to the barn be $y$ meters for each pen. The area of each pen is $A = xy=25$, so $y=\frac{25}{x}$. The total length of the fence $L$ for the four - pen setup is $L = 5x + 4y$.

Step2: Substitute $y$ into the fence - length formula

Substitute $y=\frac{25}{x}$ into $L = 5x + 4y$, we get $L(x)=5x + 4\times\frac{25}{x}=5x+\frac{100}{x}$, where $x>0$.

Step3: Find the derivative of $L(x)$

Using the power rule, if $y = ax^n$, then $y^\prime=anx^{n - 1}$. For $L(x)=5x+\frac{100}{x}=5x + 100x^{-1}$, $L^\prime(x)=5-100x^{-2}=5-\frac{100}{x^{2}}$.

Step4: Set the derivative equal to zero and solve for $x$

Set $L^\prime(x)=0$, so $5-\frac{100}{x^{2}} = 0$. Add $\frac{100}{x^{2}}$ to both sides: $5=\frac{100}{x^{2}}$. Cross - multiply to get $5x^{2}=100$. Then $x^{2}=20$, and $x = 2\sqrt{5}$ (we take the positive value since $x$ represents a length).

Step5: Find the value of $y$

Since $y=\frac{25}{x}$, substitute $x = 2\sqrt{5}$ into it. $y=\frac{25}{2\sqrt{5}}=\frac{5\sqrt{5}}{2}$.

Answer:

To minimize the amount of fence that must be used, each of the sides perpendicular to the barn should be $2\sqrt{5}$ m long and each of the sides parallel to the barn should be $\frac{5\sqrt{5}}{2}$ m long.