Answer: Maximum area = 200m
Dimensions:
Length = 10m
Width = 20m
Such that there is only one side of 20m fenced because the other side of 20m is against the rock wall.
Step-by-step explanation:
We have the next problem.
Suppose that the vegetable area is a rectangle of length L and width W.
Such that the width side coincides with the rock wall, then we have that:
The perimeter is:
P = 2*L + 2*W
But because one of the width sides coincides with the wall, the amount of fence needed will be:
P = 2*L + W = 40m
And the area can be written as:
A = W*L
We want to find the maximum area that we can have.
First, write our two equations:
40m = 2*L + W
A = W*L
First we must isolate one of the variables in the first equation, and replace it into the second equation:
40m - 2*L = W.
then:
A = (40m - 2*L)*L = 40m*L - 2*L^2
We can write this as a quadratic equation:
A(L) = -2*L^2 + 40m*L
Now, notice that it has a leading coefficient negative, then the maximum of the area function will be at the vertex of the quadratic equation.
For a quadratic equation:
f(x) = a*x^2 + b*x + c
The vertex is at:
x = -b/2a
and the point is (-b/2a, f(-b/2a))
In this case we have:
b = 40m
a = -2
Then:
we must evaluate the area function in:
L = -40m/(-2*2) = 10m
A(L) = -2*L^2 + 40m*L
A(10m) = -2*100m^2 + 400m = 200m
Now we can find the width using one of the above equations:
W = 40m - 2*L = 40m - 2*10m = 20m
Then the dimensions that the farmer should use are:
Two sides of length 10m, and one side of length 20m
