Question
a farmer wants to make a rectangular pasture with 80000 squre meters. if the pasture lies along a river and he feneces the remaning three sides,what dimension shoud he use to minimize the amount fence needed
Answers
Arora
Taking the length as x, and the width as y,
Length of fence = x + y + x = 2x + y
Now, the area has been fixed.
=> xy = k = 80,000
=> y = k/x
Function to minimize = f(x) = Length
= 2x + y
= 2x + (k/x)
f'(x) = d(2x + (k/x))/dx
= 2 - k/(x^2)
Equating f'(x) with zero,
=> 2 - k/(x^2) = 0
=> 2 = k/x^2
=> x^2 = k/2 = 80000/2 = 40000
Hence, x = +200/-200
But length cannot be negative, so x = 200
=> x = 200, y = (80000/200) = 400
Length of fence = x + y + x = 2x + y
Now, the area has been fixed.
=> xy = k = 80,000
=> y = k/x
Function to minimize = f(x) = Length
= 2x + y
= 2x + (k/x)
f'(x) = d(2x + (k/x))/dx
= 2 - k/(x^2)
Equating f'(x) with zero,
=> 2 - k/(x^2) = 0
=> 2 = k/x^2
=> x^2 = k/2 = 80000/2 = 40000
Hence, x = +200/-200
But length cannot be negative, so x = 200
=> x = 200, y = (80000/200) = 400