SOLUTION: 1. A farmer wants to make a rectangular pen along the side of a large barn and has enough material for 60 m of fencing. Only three sides must be fenced, since the barn wall will fo
Question 1159677: 1. A farmer wants to make a rectangular pen along the side of a large barn and has enough material for 60 m of fencing. Only three sides must be fenced, since the barn wall will form the fourth side. What dimensions should the farmer use so that the maximum area is enclosed? Show work. Answer by ikleyn(52781) (Show Source):
Since one side is the barn, the rectangle's fence perimeter will be
L + 2W = 60 meters.
Hence, L = 60 - 2W meters.
Area = Length * Width.
Substitute (60-2W) for L:
A = W(60 - 2W) (1)
A = -2W^2 + 60W.
It is a quadratic function. It has the maximum at x = -b/(2a), where "a" is the coefficient at the quadratic term
and "b" is the coefficient at the linear term, according to the general theory.
(See the lessons
- HOW TO complete the square to find the minimum/maximum of a quadratic function
- Briefly on finding the minimum/maximum of a quadratic function
in this site).
In your case, the maximum is at
W = = = 15.
So, W = 15 meters is the width for max area.
Then the length is L = 60 - 2W = 600 - 2*15 = 30 meters.
Find the max area. It is
A = L*W = 30*15 = 450 square meters. It is the maximum area.
