A farmer has 100 yards of fencing to form two identical rectangular pens
and a third pen that is twice as long at the other two pens, as shown in the
diagram at right. All three pens have the same width, x. What value of y produces a maximum total fenced area?
Oobleck's solution is wrong. The amount of fencing needed would actually be 5x + 6y = 100. His method to solve is still correct, though.
5x + 6y = 100
x = -(6/5)y+20
A = 2xy
A = 2y[-(6/5)y+20] = -12y^2/5 + 40y
A' = -24y/5 +40
24y = 200
y = 25/3
To solve this problem, we need to find the value of y that will maximize the total fenced area.
Let's denote the length of the two identical rectangular pens as L1 and the length of the third pen as L2. We also know that L2 is twice as long as L1. Therefore, we can say that L2 = 2L1.
Based on the information given, we know that the perimeter of each rectangular pen is equal to 100 yards.
The formula for calculating the perimeter of a rectangle is: 2*(length + width).
For the two identical rectangular pens:
Perimeter = 2(L1 + x) = 100
For the third pen:
Perimeter = 2(L2 + x) = 100
Substituting L2 = 2L1:
2(2L1 + x) = 100
Simplifying the equation, we get:
4L1 + 2x = 100
Now, let's express L1 in terms of x:
L1 = (100 - 2x)/4
L1 = (50 - x)/2
The area of each rectangular pen is equal to its length multiplied by its width.
For the two identical rectangular pens:
Area1 = L1 * x = ((50 - x)/2) * x
For the third pen:
Area2 = L2 * x = (2L1) * x = (50 - x) * x
To find the maximum total fenced area, we need to combine the areas of the two identical rectangular pens and the third pen.
Total Area = 2 * Area1 + Area2
Total Area = 2 * ((50 - x)/2) * x + (50 - x) * x
Total Area = (50 - x) * x + (50 - x) * x
Total Area = 2(50 - x) * x
To find the maximum total area, we need to take the derivative of the Total Area equation with respect to x and set it equal to 0.
d(Total Area)/dx = 2(50 - x) - 2x
0 = 2(50 - x) - 2x
0 = 100 - 4x
4x = 100
x = 25
Therefore, the value of x that produces a maximum total fenced area is 25.
Now, substituting this value of x back into the equation for L1:
L1 = (50 - x)/2
L1 = (50 - 25)/2
L1 = 25/2
L1 = 12.5
Since L2 is twice as long as L1, we can calculate L2:
L2 = 2 * L1
L2 = 2 * 12.5
L2 = 25
So, the value of y that produces a maximum total fenced area is 25, and the lengths of the two identical rectangular pens are 12.5 yards, while the length of the third pen is 25 yards.
You don't say how the pens are configured. I will assume that the long pen is adjacent to the other two, and they share a long side. So, if the pens have dimensions x*y, x*y and x*2y then the fencing needed is
5x+4y = 100
the area is
A = 2xy + x*2y = 4xy = 4x(100-5x)/4 = x(100-5x)
dA/dx = 100-10x
max area when x=10, so y = 50/4
As with all these problems, maximum area is achieved when the fencing is divided equally among lengths and widths.