Question

A farmer wants to fence a rectangular area of 800,000 m² and divide it in half with a fence that is parallel to one of the sides of the rectangle, and twice as expensive as the fence on the outer sides. How can this be done in order to minimize the cost of the fence?

Hint: If all the fencing costs the same, what are we needing to minimize in this scenario?

I gave this a try but I keep getting stuck, i would really appreciate the help!

Answer 1

xy = 800000
If the outer fence costs $c/m, then the inner fence costs $2c/m
So, the total fence cost is
x(2*c + 2c) + y*c
so we want to minimize
z = 4x + 2*800000/x
dz/dx = 4 - 1600000/x^2 = 4(x^2-400000)/x^2
we want dz/dx=0 for a minimum, so
x^2 = 400000
x = 200√10

If all the fencing costs the same, then you want to minimize
z = 3x+2y

Answer 2

I interpret this rectangel to have 2 sides (lengths) and 3 sides (widths), with one of the widths cutting the rectangle is half
let the shorter sides be x each and each of the lengths equal y
so ...
xy = 800000 ---> y = 800000/x
You want to minimize the cost, so we need a "cost" equation
I also interpret that this one dividing fence cost twice as much as the others
so assume the cost of the outer fence is 1 unit each, then
cost = 2x(1) + 2y(1) + y(2)
= 2x + 4y
= 2x + 4(800000/x)
d(cost)/dx = 2 - 3200000/x^2 = 0 for a min cost
3,200,000/x^2 = 2
x^2 = 1,600,000
x = appr 1294.9 m
y = 632.46 m

check my arithmetic

if the cost is the same for all sides, then cost = 2x(1) + 3y(1)
proceed as before.