A farmer has 400 feet of fencing with which to build a rectangular corral having two internal dividers both parallel to two of the sides of the corral. What is the maximum total area of such a corral?
I know how to maximize area, but this problem is confusing me because of the internal dividers. A step-by step explanation would help me a lot. Thanks!!
Your answer makes no sense as it confuses the variables and skips steps.
Thanks!!!
To maximize the total area of the corral, we need to find the dimensions that will use up all 400 feet of fencing while minimizing the area lost due to the internal dividers. Let's break down the problem into steps:
Step 1: Define the variables.
Let's say the length of the corral is L and the width is W. Also, let's set the length of the dividers parallel to the width as D1 and the length of the dividers parallel to the length as D2.
Step 2: Build the equation for the perimeter.
The perimeter of the corral consists of the four sides, with the two dividers subtracted. Since there are two dividers, we need to subtract them twice. The equation for the perimeter is:
2L + 2W - 2D1 - 2D2 = 400
Step 3: Rearrange the equation and solve for L.
We can rearrange the equation to solve for L:
2L = 400 - 2W + 2D1 + 2D2
L = 200 - W + D1 + D2
Step 4: Build the equation for the area.
The area of the corral is given by multiplying the length and width:
Area = L * W
Step 5: Rewrite the area equation using the value of L from step 3.
From step 3, we have:
Area = (200 - W + D1 + D2) * W
Step 6: Simplify the equation to get it in terms of one variable.
Expand the equation:
Area = 200W - W^2 + D1W + D2W
Step 7: Identify the variable that can be maximized.
To maximize the area, we can consider W as the variable and find its optimal value.
Step 8: Apply calculus to find the maximum value.
Take the derivative of the area equation with respect to W, and set it equal to zero to find the critical points.
d(Area)/dW = 200 - 2W + D1 + D2 = 0
Step 9: Solve the equation from step 8 for W.
Solve the equation:
-2W + D1 + D2 = -200
2W = D1 + D2 - 200
W = (D1 + D2) / 2 - 100
Step 10: Substitute the value of W back into the area equation.
Substitute the value of W from step 9 into the area equation from step 6:
Area = 200W - W^2 + D1W + D2W
= 200((D1 + D2) / 2 - 100) - ((D1 + D2) / 2 - 100)^2 + D1((D1 + D2) / 2 - 100) + D2((D1 + D2) / 2 - 100)
Step 11: Simplify the equation.
Expand and simplify the equation obtained in step 10 to get the maximum total area.
Once you find the maximum area, you can substitute the values of L and W back into the equation from step 3 to get the corresponding dimensions.
when it's all done, you should have the total fence divided equally among the lengths and widths.
So, with the two dividers, you have two long sides of length x, and 4 short sides of length y.
2x+4y = 400
x = 200-2y
a = xy = (200-2y)y = 200y - 2y^2
da/dy = 200-4y
da/dy=0 when y=50
so, x=100
So, 4 short lengths of 50 and 4 long sides of 100
a = xy = 100*50 = 5000