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ap calculus help please?!?
a 216m rectangular pea patch is to be enclosed by a fense and divided into 2 equal parts by another fence parallel to one of the sides. what dimensions for the outer recantgle will require the smallest total length of fence? how much fence will be needed
2 Answers
- 8 years agoFavorite Answer
3x + 2y = P
A = x * y
A = 216
216 = x * y
216/x = y
3x + 2y = P
P = 2 * (216/x) + 3x
P = 432/x + 3x
dP/dx = -432/x^2 + 3
dP/dx = 0
0 = -432/x^2 + 3
432/x^2 = 3
144/x^2 = 1
144 = x^2
12 = x
216 = 12 * y
216 / 12 = y
y = 18
12m by 18m
P = 3x + 2y
P = 3 * 12 + 2 * 18
P = 36 + 36
P = 72
72 meters of fencing
- 8 years ago
I did this same exact problem on Thursday haha. You want to optimize dimensions with smallest fencing. well you know 216=XY because X times Y is 216. F (fencing)=3X+2Y so you want only one variable and Y=216/X using the area equation. F=3X+2(216/X) to get a minimum F, take the derivative and find a relative minimum.
