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 m long and the side parallel to the barn should be m long. (type exact answers, using radicals as needed.)
Answer
Explanation:
Step1: Define variables for part a
Let the side perpendicular to the barn be $x$ and the side parallel to the barn be $y$. The perimeter equation is $2x + y=200$, so $y = 200 - 2x$. The area $A=xy=x(200 - 2x)=200x-2x^{2}$.
Step2: Find the derivative of the area function
Differentiate $A(x)=200x - 2x^{2}$ with respect to $x$. $A'(x)=\frac{d}{dx}(200x-2x^{2})=200 - 4x$.
Step3: Find the critical - points
Set $A'(x)=0$. So, $200 - 4x = 0$. Solving for $x$ gives $4x=200$, and $x = 50$.
Step4: Find the second - derivative
Differentiate $A'(x)$ to get $A''(x)=\frac{d}{dx}(200 - 4x)=-4<0$. So, when $x = 50$, the area is maximized.
Step5: Find the value of $y$
Substitute $x = 50$ into $y = 200 - 2x$, we get $y=200-2\times50 = 100$.
Answer:
The sides perpendicular to the barn should be $50$ m long and the side parallel to the barn should be $100$ m long.