Consider the following problem: A farmer with 850 ft of fencing wants to enclose

Question

Consider the following problem: A farmer with 850 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 area of the four pens? Draw several diagrams illustrating the situation, some with shallow, wide pens and some with deep, narrow pens. Find the total areas of these configurations. Does it appear that there is a maximum area? If so, estimate it. Draw a diagram illustrating the general situation. Let X denote the length of each of two sides and three dividers. Let y denote the length of the other two sides. Write an expression for the total area A in terms of both x and y. Use the given information to write an equation that relates the variables. Use part (d) to write the total area as a function of one variable. Finish solving the problem by finding the largest area.

Explanation / Answer

If we drew it out we would have 2 long sides and 5 wide segments to break it up into 4 pens.

so Perimeter = 2*L + 5*W = 850

Solve for W

W = 170 - 2/5*L

Area of each pen = L/4*W

Plug in W

Area = L/4*170 - L/4*2/5*L

Area = 42.5*L - 0.1*L^2

The maximum Area , or the derivative with respect to L = 0

0 = 42.5 - 0.2*L

L = 212.5 ft

Plug this value into the perimeter equation and solve for W

2*L + 5*W = 850

2*212.5 + 5*W = 850

5W = 850 - 425.0

5W = 425

hence W =425/5 = 85

Now,

Area for one pen = L/4*W = 212.5/4*85 = 4515.625 ft^2

hence area for all the pens = L*W = 212.5*85 = 18062.5 ft^2

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