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Easy calc problem help (10 pts for best answer)!!?
A rancher wants to fence in a rectangular area of 7000 square feet in a field and then divide the region in half with a fence down the middle parallel to one side. What is the smallest length of fencing that will be required to do this?
Could you please write out the two functions you used, at the least?
Thank you!!
3 Answers
- Some BodyLv 72 years agoFavorite Answer
Let's say w is the width of the rectangular area and h is the height. Let's assume that the fencing down the middle is parallel to the height.
The area is:
A = wh
The length of fencing is:
P = 2w + 3h
We are told that the area is 7000 square feet.
7000 = wh
w = 7000/h
P = 2 (7000/h) + 3h
P = 14,000 / h + 3h
We want P to be a minimum, so find dP/dh and set to 0:
dP/dh = -14,000 / h² + 3
0 = -14,000 / h² + 3
14,000 / h² = 3
h² = 14,000 / 3
h ≈ 68.3 ft
w ≈ 102.5 ft
P ≈ 409.9 ft
- MichaelLv 72 years ago
This is the second time you've asked this.
See my previous answer:
410 ft <–––––
- ?Lv 62 years ago
A = L * W
7,000 = L * W
L = 7,000 / W
P = 2L + 3W
P = 14,000 / W + 3W
P = 14,000 W^-1 + 3W
dP/dW = 0 = -14,000 W^-2 + 3
14,000 = 3 W^2
W = (14,000 / 3)^0.5
W = 68.3 ft.
L = 102.5 ft.
P = 3 (68.3) + 2 (102.5)
P = 410 ft. (to nearest foot)
