Problem

A \(\text{three}\text{-}\text{sided}\) rectangular pen uses \(60~\text{ft}\) of fencing against a wall. Let \(x\) be each perpendicular width and \(y\) the side parallel to the wall. Use the diagram to build a \(\text{one}\text{-}\text{variable}\) area model with its positive domain, then find the maximizing dimensions and maximum area.

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From 2x + y = 60, y = 60 - 2x and A(x) = x(60 - 2x), with 0 < x < 30. The maximum occurs at x = 15 ft and y = 30 ft, giving 450 ft².

From x + y = 60, y = 60 - x and A(x) = x(60 - x), with a 30-ft by 30-ft maximum; this omits one of the two fenced widths.

Use y = 60 - 2x but set area equal to x + y, so the maximizing dimensions are chosen from a length sum instead of the product xy.

Use y = 60 - 2x and A(x) = x(60 - 2x), but include x = 0 and choose the 0-by-60 degenerate pen as the maximum.

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