Lester wants to build a rectangular enclosure for his animals. One side of the pen will be against the barn, so he needs no fence on that side. The other three sides will be enclosed with wire fencing. If Lester has 850 feet of fencing, you can find the dimensions that maximize the area of the enclosure. a. Let $w$ be the width of the enclosure (perpendicular to the barn) and let $l$ be the length of the enclosure (parallel to the barn). Write a function for the area $A$ of the enclosure in terms of $w$. \[ A(w)= \] b. What width $w$ would maximize the area? \[ w= \] $\mathrm{ft}$ c. What is the maximum area? \[ A= \] square feet

Step 1 ：Let's denote the width of the enclosure as \(w\) and the length of the enclosure as \(l\). Since one side of the enclosure will be against the barn, we only need to use the fencing for the other three sides. This means that the total length of the fence, which is 850 feet, is equal to twice the width plus the length (\(2w + l = 850\)).

Step 2 ：We can express the length \(l\) in terms of the width \(w\) by rearranging the equation: \(l = 850 - 2w\).

Step 3 ：The area \(A\) of the rectangle is given by the product of the width and the length, so we can express the area in terms of \(w\) as follows: \(A(w) = w * (850 - 2w)\).

Step 4 ：To find the width that maximizes the area, we need to find the derivative of the area function and set it equal to zero. The derivative of \(A(w)\) is \(850 - 4w\).

Step 5 ：Solving the equation \(850 - 4w = 0\) gives us the critical points. The only critical point in this case is \(w = 425/2\).

Step 6 ：Substituting \(w = 425/2\) into the area function gives us the maximum area: \(A = 180625/2\).

Step 7 ：Final Answer: The width that would maximize the area is \(\boxed{212.5}\) feet and the maximum area is \(\boxed{90312.5}\) square feet.