Asked by Gwapo
A farmer wants to fence in an area of 15000 m² in a rectangular field and then divide it into half with a fence parallel to one sides of the rectangle. How can he do this so as to minimize the cost of the fence?
Answers
Answered by
PsyDAG
Are you assuming 150 by 100? If so, the amount of fence would be 3(100)+2(150). Compare that to other possibilities, e.g., 300 by 50.
Answered by
Reiny
Let the long side by y m
let the shorter side be x m
So the area is xy and xy = 15000, -----> y = 15000/x
For the cost to be minimum , the total amount of fencing has to be minimum
F = 2y + 3x
= 2(15000/x) + 3x
dF/dx = -30000/x^2 + 3
= 0 for a minimum of F
3 = 30000/x^2
x^2 = 10000
x = 100 , then y = 15000/x = 150
State your conclusion
let the shorter side be x m
So the area is xy and xy = 15000, -----> y = 15000/x
For the cost to be minimum , the total amount of fencing has to be minimum
F = 2y + 3x
= 2(15000/x) + 3x
dF/dx = -30000/x^2 + 3
= 0 for a minimum of F
3 = 30000/x^2
x^2 = 10000
x = 100 , then y = 15000/x = 150
State your conclusion