A farmer has 100 yards of fencing to form two identical rectangular pens and a third pen that is twice as long as the other two pens, as shown in the diagram to the right. All three pens have the same width, x. Which value of y produces the maximum total fenced area?
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We do not have your diagram, so we don't know if they have common fences.
To find the value of y that produces the maximum total fenced area, we need to optimize the area of the three pens.
Let's start by labeling the sides of the pens:
- The two identical rectangular pens have length y and width x.
- The larger pen has length 2y and width x.
Now, let's calculate the total length of fencing required:
For the two identical rectangular pens:
- Each pen has two sides of length y and two sides of length x.
- So the total length of fencing required for the two identical pens is: 2y + 2x.
For the larger pen:
- The large pen has two sides of length 2y and two sides of length x.
- So the total length of fencing required for the large pen is: 2(2y) + 2x = 4y + 2x.
Given that we have 100 yards of fencing, we can set up the equation:
2y + 2x + 4y + 2x = 100
Simplifying the equation:
6y + 4x = 100
3y + 2x = 50
Now, let's express y in terms of x:
3y = 50 - 2x
y = (50 - 2x)/3
To find the maximum total fenced area, we need to maximize the area of the three pens. The area of each rectangle is given by the formula length times width.
The area of the two identical rectangular pens is: A1 = y * x = (50 - 2x)x/3 = (50x - 2x^2)/3
The area of the larger pen is: A2 = 2y * x = 2(50 - 2x)x/3 = (100x - 4x^2)/3
Finally, the total fenced area is given by the sum of the individual areas:
Total Area, A = A1 + A1 + A2 = (50x - 2x^2)/3 + (50x - 2x^2)/3 + (100x - 4x^2)/3
= (200x - 8x^2)/3
To find the value of y that produces the maximum total fenced area, we need to find the maximum value of A. This occurs when the derivative of A with respect to x is equal to zero.
Taking the derivative of A with respect to x:
dA/dx = (200 - 16x)/3
Setting the derivative equal to zero:
(200 - 16x)/3 = 0
200 - 16x = 0
16x = 200
x = 200/16
x = 12.5
Now, substitute this value of x back into the equation for y:
y = (50 - 2x)/3 = (50 - 2(12.5))/3 = (50 - 25)/3 = 25/3
So, the value of y that produces the maximum total fenced area is y = 25/3.