Solve an optimization problem for the maximum area rectangle under a constraint

Primary Tool

Math Workspace (CAS)

CAS is useful here because it keeps the algebraic setup, the derivative test, and the final numerical area check in one place. The graph then shows why the critical point is the top of a downward-opening parabola.

Problem

A rectangle has perimeter $24$ . What dimensions maximize its area?

The point of the problem is that one geometric condition produces a one-variable objective function, so the derivative test answers a question about actual dimensions.

How the constraint reduces the problem

If the side lengths are $x$ and $y$ , then the perimeter condition gives $2 x + 2 y = 24$ , so $y = 12 - x$ .

That substitution removes one variable. The area becomes $A (x) = x (12 - x) = 12 x - x^{2}$ , so maximizing the rectangle area is the same as maximizing a quadratic function on the feasible interval.

Step-by-step walkthrough

The important move is replacing the geometric constraint with a one-variable area function that can be differentiated directly.

1Write the perimeter constraint as $2 x + 2 y = 24$ and solve for one side: $y = 12 - x$ .
2Substitute into the area formula $A = x y$ to get $A (x) = x (12 - x) = 12 x - x^{2}$ .
3Differentiate to find $A^{'} (x) = 12 - 2 x$ .
4Solve $A^{'} (x) = 0$ to find the critical point $x = 6$ .
5Differentiate again to get $A^{''} (x) = - 2$ , so the graph is concave down and the critical point gives a maximum.
6Compute the matching side length $y = 12 - 6 = 6$ and the maximum area $A (6) = 36$ .

Common Pitfall

The derivative is not taken on the original two-variable area formula by itself. You first have to use the constraint to rewrite the area in one variable, otherwise the optimization step is incomplete.

Try a Variation

Redo the same problem with perimeter $40$ . Which formulas change, and why does the maximizing rectangle still turn out to be a square?

Keep moving through the cluster

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