A rancher has 1000 m of fencing to enclose two rectangular corrals. The corrals have the same dimensions and one side in common. What dimensions will maximize the enclosed area?
2007-04-29 13:27:11 · 1 answers · asked by slow_math 1 in Science & Mathematics ➔ Mathematics
If h is the height of the corrals (including the shared side) and w is the width...
The perimeter is 3h + 4w = 1000
Solved for w: w = 250 - 3h/4
You are trying to maximize 2wh, the area of the pair of corrals.
Substituting the equation solved for w, you are maximizing:
2wh =
2(250-3h/4)(h) =
500h - 3h^2/2
To maximize, take the derivative and find where it is zero:
d(500h - 3h^2/2)/dh = 0
500 - 3h = 0
h = 500/3
So you would be constructing corrals of 500/3 meters height (3 sides = 500m) and 125 meters width (4 sides = the other 500m)
The total area of the two would be:
500/3 (height) * 125 (width) * 2 (corrals) =
125,000/3 m^2
~41,667 m^2
2007-04-29 13:50:31 · answer #1 · answered by McFate 7 · 1⤊ 0⤋