A rancher wants a fence in an area?

Question

A rancher wants to fence in an area of 400 square feet in a rectangular field using fencing material costing 1.2 dollars per foot, and then divide it in half down the middle with a partition, parallel to one side, constructed from material costing 0.5 dollars per foot.
Assuming that the partition is parallel to the side which gives the width of the field, find the dimensions of the field of the cheapest design.
What is the length?
What is the width?
What is the (total) cost of the cheapest design?

Answer 1

let w = width
length = 400/w
perimeter = 2w + 800/w
cost = 1.2*(2w + 800/w) + 0.5*w
= 2.9w + 960/w
d/dw = 2.9 - 960/w^2 = 0 for min
w^2 = 960/2.9
w = 18.19, L = 21.98, cost = $105.53

Answer 2

Hello,

Let l = the length
let w = the width

Now the area = 400 = L*W so l = 400/w Then

The perimeter is 2w + 2*(400/w) then the cost at 1.2 is
c = 1.2*(2w + 2*(400/w)) plus the divider which is 0.5w so

c = 1.2*(2w + 2 *(400/w) + 0.5w

Simplify and we have 2.4w + 960/w +0.5w = 2.9w + 960/w now take the derivative and we obatin

c' = 2.9- 960/w^2

Now to minimize we set it equal to 0 so

0 = 2.9 - 960/w^2 so 2.9w^2 - 960 = 0 then w^2 = 331.03 then
w = 18.19

l = 400/18.19 = 21.99

Now put these into the cost equation to find the cheapest cost.


Hope This Helps!!!