Maya had been wrestling with the problem all semester. It was the sort of question that seemed simple at first glance, then revealed hidden layers like an onion. The statement asked her to , using only one variable. In other words, the box’s height and the side of its base were tied together by the geometry of the sphere, and the challenge was to express the volume in terms of a single unknown, then locate its critical point.
Maya wrote the result in bold, underlined it, and added a small smiley face next to it—her personal signature of triumph. The next morning, the professor walked into the seminar room, a stack of papers in his hand. He asked the class to volunteer a solution for Exercise 179. Maya’s hand rose, heart thudding like a metronome.
She pulled a chair, settled into the worn leather, and spread out her notes. The room was quiet except for the distant hum of the campus heating system and the occasional rustle of a late‑night janitor’s cart. Maya began by sketching the situation on a scrap of graph paper. A sphere centered at the origin, radius R , and a rectangular box whose center coincided with the sphere’s center. Because the base was a square, she let x denote the length of one side of the base, and y the height of the box.
Factoring out the common denominator gave Maya had been wrestling with the problem all semester
When the old brass bell of the university’s clock tower struck eleven, Maya slipped the final key into the lock of the library’s rare‑books room. The room smelled of polished oak, leather, and a faint hint of coffee—its only occupants the towering shelves that held the most beloved (and most feared) tomes of the mathematics department.
Discarding the trivial solution (x = 0) (which gave zero volume), she solved
Finally, the maximal volume:
She felt a surge of satisfaction. The problem had been reduced to a single‑variable function, exactly as the title promised. The next step was to find the maximum of (V(x)). Maya knew she needed the derivative (V'(x)) and the critical points where it vanished (or where the derivative was undefined). She set her mind to the task.
The vertices of the box lie on the sphere, so each corner satisfies the equation
Using the product rule and the chain rule, she obtained In other words, the box’s height and the
As she walked home, she imagined the inscribed cube—edges perfectly aligned, each corner just touching the sphere—sitting like a gem inside a glass sphere, a concrete reminder that sometimes, the most beautiful solutions are the simplest, and that every calculus problem hides a story waiting to be told.
Plugging this back into the expression for :
[ V(x) = x^2 \cdot y = x^2 \cdot 2\sqrt{R^2 - \frac{x^2}{2}} = 2x^2\sqrt{R^2 - \frac{x^2}{2}} . ] He asked the class to volunteer a solution for Exercise 179