Manuel wants to build a rectangular pen for his animals. One side of the pen will be against the barn; the other three sides will be enclosed with wire fencing. If Manuel has 750 feet of fencing, what dimensions would maximize the area of the pen?
a) Let w be the length of the pen perpendicular to the barn. Write an equation to model the area of the pen in terms of w
-----I think this part should be w^2 but that isn't the correct answer------
b) What width w would maximize the area?
-----i know squares have maximized area
But it seems like I keep solving this problem and getting the wrong answer.-----
You only have three sides, usual solution does not apply
2 w + L = 750 (note not 2 L)
so L = 750 - 2 w
A = w L = w (750-2w)
-2w^2 + 750 w = A
w^2 - 375 w = -A/2
w^2 - 375 w + (375/2)^2 = -A/2 + 35156
(w-375/2)^2 = -(1/2)(A -70,312)
max at w = 375/2 = 188
a) Let's denote the width of the pen perpendicular to the barn as w. Since it is stated that the pen is rectangular, the length parallel to the barn can be denoted as l. The area of the pen can be modeled as the product of the length and width, so the equation would be:
Area = l * w
However, we need to express l in terms of w. Since one side is against the barn, the length l will be equal to the remaining two sides. For these two sides, we would have:
2w + l = 750
Therefore, we can rearrange the equation to solve for l:
l = 750 - 2w
Now we can substitute this expression for l into our area equation:
Area = w * (750 - 2w)
b) To maximize the area, we need to find the value of w that yields the maximum value for the area. One way to find this is by taking the derivative of the area equation with respect to w and setting it equal to zero:
d(Area)/dw = 750 - 4w
Setting this equal to zero and solving for w:
750 - 4w = 0
4w = 750
w = 750/4
w = 187.5
So, the width that would maximize the area is 187.5 feet.
To solve this problem, let's go step by step:
a) To model the area of the pen in terms of the width (w), we can use the formula for the area of a rectangle: Area = length × width. The length of the pen perpendicular to the barn will be 750 - 2w (assuming the barn takes up one side of the pen). So the equation will be:
Area = (750 - 2w) × w
b) To find the width (w) that maximizes the area, we need to differentiate the equation we obtained in part a) with respect to w, and then find the value of w where the derivative equals 0.
Taking the derivative of the equation with respect to w:
d(Area)/dw = 750 - 4w
Setting it equal to 0 to find the critical point:
750 - 4w = 0
4w = 750
w = 750/4
w = 187.5
So, the width (w) that maximizes the area of the pen is 187.5 feet.
Note: It's important to remember that the width cannot be negative or greater than 375 (half of the total length of fencing).