Suppose you wish to enclose a rectangular area with a divider using 600 feet of fence. What dimensions should you use in- order to enclose the most possible area?
2006-11-25 16:25:01 · 6 answers · asked by Eric 2 in Science & Mathematics ➔ Mathematics
In general, to maximize the area enclosed in a rectangle, you should use a square area. So the length and width should be equal, and they should each be equal to 600/4 or 150 feet.
If you need to prove this mathematically, you have to start by formulating an equation for the area. (I'll let you know now that this involves calculus, so if you're not there yet, don't worry about the rest of this.)
A = LW, and 2L + 2W must equal 600, so L + W = 300 and L = 300 - W. A = W(300 - W) = 300W - W^2. That's your equation. Now you have to maximize it. This is the calculus part. You have to take the derivative of that equation, which is 300 - 2W, and set it equal to 0. That means 300 - 2W = 0: 300 = 2W: W = 150. At that point, the area is at a maximum. (Mathematically, it could also be a minimum, but in this case it's a maximum.) Therefore the width is 150, and so is the length.
Edit: noviadluisme, a square is a rectangle. It's a special type of rectangle.
2006-11-25 16:34:23 · answer #1 · answered by Amy F 5 · 0⤊ 0⤋
Remember there has to be a divider, so we have
600 = 3W + 2L
A = W*L = [200 - (2/3)L] * L = 200L - (2/3)L^2
A' = 200 - (4/3) L
0 = 200 - (4/3) L
L = 150
W = 100
2006-11-25 16:41:19 · answer #2 · answered by z_o_r_r_o 6 · 0⤊ 0⤋
if u put a fence of 298 m x 1m, it means the area is 298 m2. lower it and ull see that when u get close to the square ur area increases..150 x150 = 22500 m2
2006-11-25 16:35:23 · answer #3 · answered by iidibitizi 3 · 0⤊ 0⤋
A square, 150 feet long on each side.
Doug
2006-11-25 16:32:02 · answer #4 · answered by doug_donaghue 7 · 0⤊ 0⤋
150 feet for one side... the area is a square
2006-11-25 16:29:45 · answer #5 · answered by arumisan 2 · 1⤊ 0⤋
150' square.
2006-11-25 17:11:23 · answer #6 · answered by yupchagee 7 · 0⤊ 0⤋