SOLUTION: A farmer has 336 running feet of fence and wishes to form a rectangular pasture. One side of the pasture will be bounded by a very long wall, and so no fencing material will be ne

Algebra ->  Surface-area -> SOLUTION: A farmer has 336 running feet of fence and wishes to form a rectangular pasture. One side of the pasture will be bounded by a very long wall, and so no fencing material will be ne      Log On


   

Question 192071: A farmer has 336 running feet of fence and wishes to form a rectangular pasture. One side of the pasture will be bounded by a very long wall, and so no fencing material will be needed for that side of the pasture. What should the dimensions of the pasture be if the area of the pasture is to be largest possible?

Could you please provide a detailed explanation of how you work the problem to find the solution. Thanks!
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
A farmer has 336 running feet of fence and wishes to form a rectangular pasture. One side of the pasture will be bounded by a very long wall, and so no fencing material will be needed for that side of the pasture. What should the dimensions of the pasture be if the area of the pasture is to be largest possible?
.
Let x = width
and y = length
.
From the provided length of fencing available:
2x+y = 336
Solving for y:
y = 336-2x
.
Area = xy
Plugging in our value of y:
Area = x(336-2x)
Area = 336x-2x^2
Area = -2x^2 + 336x (parabola that opens downwards)
.
If we find the vertex, we'll find the maximum value of the width:
Area = -2x^2 + 336x
Area = -2(x^2 - 168x)
Area = -2(x^2 - 168x + 7056) + 14112
Area = -2(x - 84)^2 + 14112
Since the vertex is at (84, 14112)
we know the width is 84 feet
.
Length:
2x+y = 336
2(84)+y = 336
168+y = 336
y = 168 feet (length)
.
Dimension:
84 feet by 168 feet