Fence on the Farm

[charlie33]

A farmer wants to enclose a rectangular field along a river on three sides. If 4,400 feet of fencing is to be used, what dimensions will maximize the enclosed area?

A. 1,100 ft (parallel to the river) by 1,100 ft
B. 1,100 ft (parallel to the river) by 2,200 ft
C. 2,200 ft (parallel to the river) by 1,100 ft
D. 2,200 ft (parallel to the river) by 2,200 ft
E. 4,400 ft (parallel to the river) by 1,100 ft

[GMATQuantCoach]

First, you can't use more than 4400 ft of fencing. This alone eliminates B,D,and E. To maximize the area, most likely you will need to use all 4400ft of fencing. Then C is your answer.

[dtweah]

A farmer wants to enclose a rectangular field along a river on three sides. If 4,400 feet of fencing is to be used, what dimensions will maximize the enclosed area?

A. 1,100 ft (parallel to the river) by 1,100 ft
B. 1,100 ft (parallel to the river) by 2,200 ft
C. 2,200 ft (parallel to the river) by 1,100 ft
D. 2,200 ft (parallel to the river) by 2,200 ft
E. 4,400 ft (parallel to the river) by 1,100 ft

2X +Y=4400

Only C meets the above condition, where Y is length || to river

[truplayer256]

2Y+X=4400

Y=Width of rectangular fence
X=Length of rectangular fence

Area=XY

(4400-2Y)(Y)<--- we want to maximize this.

4400Y-2Y^2

take the first derivative of the above equation and set it equal to 0.

4400-4Y=0
Y=1,100
X=4400-2200=2200

C