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.
After the class, several classmates gathered around Maya, peppering her with questions. She explained how the symmetry of the sphere forced the optimal box to be a cube, and how the derivative’s denominator reminded her to stay within the physically meaningful interval (0 < x < \sqrt{2},R). Later that night, Maya returned the Thomas & Finney volume to its shelf, the thin solution sheet now neatly folded back into place. She closed the library’s heavy door and stepped into the cool campus air, the bell of the clock tower echoing the rhythm of her thoughts. She felt a surge of satisfaction
[ y = 2\sqrt{R^2 - \frac{1}{2}\Bigl(\frac{2R}{\sqrt{3}}\Bigr)^2} = 2\sqrt{R^2 - \frac{1}{2}\cdot\frac{4R^2}{3}} = 2\sqrt{R^2 - \frac{2R^2}{3}} = 2\sqrt{\frac{R^2}{3}} = \frac{2R}{\sqrt{3}}. ] Maya knew she needed the derivative (V'(x)) and
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. She explained how the symmetry of the sphere
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.
[ y = 2\sqrt{R^2 - \frac{x^2}{2}} . ]
[ V_{\max}= x^2 y = \Bigl(\frac{2R}{\sqrt{3}}\Bigr)^2 \cdot \frac{2R}{\sqrt{3}} = \frac{4R^2}{3} \cdot \frac{2R}{\sqrt{3}} = \frac{8R^3}{3\sqrt{3}}. ]