Assuming a rectangular fenced area, let the area measure
feet by
feet. Then the length of fencing used is such that
![2x + y = 900 \implies y = 900 - 2x](https://img.qammunity.org/2023/formulas/mathematics/college/bzx5g6fo812wu0tjal8r5w9m4s7ryny7dc.png)
where
is the length of the side parallel to the existing wall.
The area enclosed by the fence is
![A(x,y) = xy \implies A(x) = x(900-2x) = 900x - 2x^2](https://img.qammunity.org/2023/formulas/mathematics/college/9l2simjq79a2lpo4z8biknljojn6aq9h2d.png)
Find the critical points of
.
![A'(x) = 900 - 4x = 0 \implies x = 225](https://img.qammunity.org/2023/formulas/mathematics/college/ll94akt6gl637cdzeforhcwrmvm7mlunop.png)
If two sides measure 225 feet, then the remaining side must measure
![y = 900 - 2\cdot225 = 450](https://img.qammunity.org/2023/formulas/mathematics/college/lee20ag4g7wiooqb0htvx4l6dew5o4oy5r.png)
feet. So the pen must have dimensions 225-ft by 450-ft. We know this gives a maximum area, since by completing the square we find an upper bound for the enclosed area of
![900x - 2x^2 = -2 (x^2 - 450x) \\\\ ~~~~~~~~ = -2 (x^2 - 450x + 225^2 - 225^2) \\\\ ~~~~~~~~ = 2\cdot225^2 - 2 (x - 225)^2 \\\\ ~~~~~~~~ \le 2\cdot225^2 = 101,250](https://img.qammunity.org/2023/formulas/mathematics/college/zs1ja1g3a1fq4o7al52p43azftebp8y83u.png)
square feet.