Mathematics

Mathematics

Problem

An open-top box with a square base is to be constructed from 200 square inches of cardboard. If the base has side length ss inches, which value of ss maximizes the volume of the box? a) 1003\sqrt{\dfrac{100}{3}} b) 2003\sqrt{\dfrac{200}{3}} c) 525\sqrt{2} d) 503\sqrt{\dfrac{50}{3}}
Intuitive explanation

The key is to express volume as a function of a single variable, then optimize it. Since you have a fixed amount of material (200 sq in), the base area and four side walls compete for that cardboard—as the base gets bigger, the walls get shorter. You'll set up a constraint equation from the material limit, solve for height in terms of ss, substitute into the volume formula V=s2hV = s^2 h, then find the maximum using calculus or algebra.

Intuitive explanation

The key is to express volume as a function of a single variable, then optimize it. Since you have a fixed amount of material (200 sq in), the base area and four side walls compete for that cardboard—as the base gets bigger, the walls get shorter. You'll set up a constraint equation from the material limit, solve for height in terms of ss, substitute into the volume formula V=s2hV = s^2 h, then find the maximum using calculus or algebra.

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