Consider the following problem: A farmer with 810 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total are of the four pens?
your drawing should look like this
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then let the width of the smaller rectangle be x, let its length be y
so the amount of fencing would be
8x + 5y
so 8x+5y=810
y=(810-8x)/5
Area = 4xy = 4x(810-8x)/5
= 648x - 32x^2 /5
d(Area)/dx = 648 - (64/5)x = 0 for a max of Area
solve to get x=50.625
and subst to get y=81
for a max area of 4(50.625)(81)=16402.5
sorry, my diagram did not come out the way I hoped it would.
It is a large rectangle cut into four smaller ones.
Thank you!!!!!!!!!!
To solve this problem, we want to find the dimensions of the rectangle that would maximize the total area of the four pens.
First, let's label the width of the smaller rectangle as x and its length as y.
The total amount of fencing would be the perimeter of the small rectangle plus the dividing fences. The perimeter of the small rectangle is 2x + 2y, and we need two dividing fences of length x and two dividing fences of length y. So, the total length of the fences is:
Fencing = 2x + 2y + x + y + x + y = 4x + 5y
We are told that the total length of the fencing is 810 ft, so we can write the equation:
4x + 5y = 810
Next, let's express y in terms of x by rearranging the equation:
5y = 810 - 4x
y = (810 - 4x) / 5
Now, let's find the area of the four pens. The area of one pen is simply x * y, so the total area of the four pens is 4xy:
Area = 4xy
Substituting the expression for y:
Area = 4x * (810 - 4x) / 5
To find the maximum area, we need to find the critical points where the derivative of the area with respect to x equals zero. So, let's differentiate the expression for the area with respect to x:
d(Area) / dx = 4 * [810 - 4x] / 5 - (64/5) * x
Setting the derivative equal to zero:
4 * [810 - 4x] / 5 - (64/5) * x = 0
Simplifying the equation:
648 - (64/5)x = 0
From here, we solve for x:
648 - (64/5)x = 0
(64/5)x = 648
x = (648 * 5) / 64
x = 50.625
Substituting x back into the equation for y:
y = (810 - 4 * 50.625) / 5
y = 81
Therefore, the dimensions of the rectangle that maximize the total area of the four pens are approximately 50.625 ft for the width (x) and 81 ft for the length (y).
Finally, we can calculate the maximum area by plugging these values back into the area equation:
Area = 4 * 50.625 * 81
Area ≈ 16402.5 square ft
So, the largest possible total area of the four pens is approximately 16402.5 square ft.