[ 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}} . ]
First, she rewrote the volume in a friendlier form for differentiation:
Maya solved for in terms of x :
[ 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}}. ]
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. [ V(x) = x^2 \cdot y = x^2
[ 4xR^2 - 3x^3 = 0 \quad\Longrightarrow\quad x\bigl(4R^2 - 3x^2\bigr) = 0. ]
Factoring out the common denominator gave The room was quiet except for the distant
[ V'(x) = \frac{4x\bigl(R^2 - \tfrac{x^2}{2}\bigr) - x^3}{\sqrt{R^2 - \tfrac{x^2}{2}}} = \frac{4xR^2 - 2x^3 - x^3}{\sqrt{R^2 - \tfrac{x^2}{2}}} = \frac{4xR^2 - 3x^3}{\sqrt{R^2 - \tfrac{x^2}{2}}}. ]