Problem

A lidless box has square base side \(x\), height \(h\), and material area \(A~>~0\). Use the diagram and constraint to build a \(\text{one}\text{-}\text{variable}\) volume model with its positive interval, then find the optimizer, dimensions, and maximum volume.

Prompt lidless box with square base side x, height h, one base, four side faces, and total material area A; the eliminated model, interval, optimizer, dimensions, and maximum are withheld.

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Use x² + 2xh = A, omitting two of the four side faces, and optimize that incomplete material-area model.

Use the correct formula but include zero-height or zero-base endpoints and choose a degenerate zero-volume box as the maximum.

From x² + 4xh = A, h = (A - x²)/(4x) and V(x) = x(A - x²)/4, with 0 < x < sqrt(A). The maximum occurs at x = sqrt(A/3), where h = x/2 and Vmax = A³/²/(6sqrt(3)).

Leave the base side and height independent and conclude that no one-variable model or optimizer is determined, despite the stated square base and material constraint.

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A lidless box has square base side \(x\), height \(h\), and material area \(A~>~0\). Use the diagram and constraint to build a \(\text{one}\text{-}\text{variable}\) volume model with its positive interval, then find the optimizer, dimensions, and maximum volume.

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