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an ostrich farmer wants to enclose a rectangular area and then divide it into 6 pens with fencing parallel to one side of the rectangle. there are 730 feet of fencing available to complete the job. what is the largest possible total area of the 6 pens?

User Fxbois
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Let's call the length of the rectangle L and the width W. We can set up an equation for the total amount of fencing required:

2L + 7W = 730

We need 2L because there are two lengths of the rectangle that need to be fenced, and 7W because we are dividing the rectangle into 6 pens, so we need 7 widths of fencing (one between each pair of pens and two on the ends).

Solving for L, we get:

L = (730 - 7W) / 2

The total area of the 6 pens is given by:

A = 6(L/6)(W)

where (L/6) is the length of each pen and W is the width. Simplifying this expression, we get:

A = (LW)/3

Now we can substitute for L in terms of W:

A = [(730 - 7W)W]/6

Expanding and simplifying, we get:

A = (730W - 7W^2)/6

To find the maximum possible area, we can take the derivative of this expression with respect to W, set it equal to zero, and solve for W:

dA/dW = (730 - 14W)/6 = 0

730 - 14W = 0

W = 52.14

We can plug this value back into the equation for L to find the dimensions of the rectangle:

L = (730 - 7W) / 2 = 312.43

Therefore, the largest possible total area of the 6 pens is:

A = (730W - 7W^2)/6 = 13640.11 square feet (rounded to two decimal places).

Answer: 13640.11 square feet.
User Antoniom
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