A rectangular area is to be enclosed and divided into thirds. The family has $700 to spend for the fencing material. The outside fence costs $12 per running foot installed, and the dividers cost $30 per running foot installed. What are the dimensions that will maximize the area enclosed?
Draw a picture of the problem. You should get to the forumlae 700=24L+84W with basic math. Solve for L (or W). A=LxW. Maximum Area occurs when derivative = 0. Take derivative and solve for L (or W).
To find the dimensions that will maximize the area enclosed, we need to determine the length and width of the rectangular area.
Let's assume the length of the rectangular area is L and the width is W.
The outside fence consists of two lengths and two widths of the rectangle, totaling a perimeter of 2L + 2W. The cost of the outside fence will be $12 per running foot, so the cost of the outside fence will be 12 * (2L + 2W).
The dividers divide the rectangular area into thirds. Since the area is divided into thirds vertically, there will be two dividers of length W, and since the area is divided into thirds horizontally, there will be two dividers of length L. The total length of the dividers is 2W + 2L.
Therefore, the cost of the dividers will be $30 per running foot, so the cost of the dividers will be 30 * (2W + 2L).
The total cost of the fencing material is the sum of the cost of the outside fence and the cost of the dividers. So,
Total cost of fencing material = cost of outside fence + cost of dividers
= 12 * (2L + 2W) + 30 * (2W + 2L)
= 12 * 2L + 12 * 2W + 30 * 2W + 30 * 2L
= 24L + 24W + 60W + 60L
= 84L + 84W
Since the family has $700 to spend on fencing material, we can set up the equation:
84L + 84W = 700
Now we need to consider the area of the rectangular area. The area of the rectangle is given by A = L * W.
To maximize the area, we can express the area in terms of a single variable and then use calculus.
We have the equation 84L + 84W = 700, so we can solve it for L:
84L = 700 - 84W
L = (700 - 84W) / 84
Substituting this value of L into the area equation, we get:
A = (700 - 84W) / 84 * W
= (700W - 84W^2) / 84
Now, we can take the derivative of A with respect to W and set it equal to zero to find the maximum:
dA/dW = (700 - 168W) / 84
Setting dA/dW = 0, we get:
700 - 168W = 0
168W = 700
W = 700/168
W = 25/6
Substituting this value of W back into the earlier equation to solve for L, we get:
L = (700 - 84W) / 84
L = (700 - 84 * (25/6)) / 84
L = (700 - 7 * 25) / 6
L = 425/6
Therefore, the dimensions that will maximize the area enclosed are L = 425/6 and W = 25/6.