Problem

A rectangle has perimeter \(40~\text{units}\), and \(\text{one}~\text{side}\) is \(x\). Use the diagram to build a \(\text{one}\text{-}\text{variable}\) area model with its positive feasible domain, then find the maximizing dimensions and maximum area.

Prompt rectangle with sides x and y and fixed perimeter 40, so 2x+2y=40; the area model, feasible domain, optimizer, dimensions, and maximum area are withheld.

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What this problem is really about

Turn the fixed perimeter into a relation between the two side lengths, counting both copies of each side. Solve that constraint for one dimension and substitute into length times width to create a one-variable quadratic. Restrict the variable so both dimensions stay positive, then use the downward-opening quadratic's vertex to locate the feasible maximum and recover the other side.

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From x + y = 40, y = 40 - x and A(x) = x(40 - x), with a 20-by-20 maximum; this treats the two side lengths as the entire perimeter.

Use y = 20 - x and A(x) = x(20 - x), but include x = 0 and choose the degenerate 0-by-20 rectangle as the maximum.

Use y = 20 - x but set area equal to x + y = 20, so every positive x is claimed to give the same maximum area.

From 2x + 2y = 40, y = 20 - x and A(x) = x(20 - x), with 0 < x < 20. The maximum occurs at x = 10, giving a 10-by-10 rectangle with area 100 square units.

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A rectangle has perimeter \(40~\text{units}\), and \(\text{one}~\text{side}\) is \(x\). Use the diagram to build a \(\text{one}\text{-}\text{variable}\) area model with its positive feasible domain, then find the maximizing dimensions and maximum area.

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