Question 1163322
this turned my head around a few times, but i think i have it.
the corral forms two rectangles that are joined together on one side.
the total fencing available is 120 meters.
the total fencing will be 2 * the length of the enclosure plus 3 times the width of the enclosure.
the area of the enclosure is the length * width.
let:
L = length
W = width
A = area
F = fence
your equations are:
F = 2L + 3W
A = L * W
since F = 120, you get:
120 = 2L + 3W
A = L * W
from 120 = 2L + 3W, solve for L to get:
L = (120 - 3W) / 2
from A = L * W, replace L with (120 - 3W) / 2 to get:
A = (120 - 3W) / 2 * W
simplify to get:
A = 60 * W - 1.5 * W^2
rearrange the terms on the right side of thise equation by descendng order of degree to get:
A = -1.5 * W^2 + 60 * W
this is a quadratic equation in standard form where:
a = -1.5
b = 60
maximum value will be when W = -b/2a = 3/60 = 20
when W = 20, maximum area will be -1.5 * 20^2 + 60 * 20 = 600 square meters.
since L * W = A, then L = A / W = 600 / 20 = 30
you get:
L = 30
W = 20
F = 2L + 3W becomes F = 2*30 + 3*20 = 60 + 60 = 120, so this part checks out.
your solution is that the maximum area will be 600 square meters.
note that the perimeter is equal to 2L + 2W, making the perimeter equal to 60 + 40 = 100 meters.
you can graph the equation of area = -1.5 * W^2 + 60 * W by making area equal to y and W equal to x.
the equation becomes y = -1.5 * x^2 + 60 * x
that looks like this on the graph.


<img src = "http://theo.x10hosting.com/2020/081201.jpg" >