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.