Finally, the maximal volume:
Discarding the trivial solution (x = 0) (which gave zero volume), she solved
which simplified to
Using the product rule and the chain rule, she obtained
[ 4xR^2 - 3x^3 = 0 \quad\Longrightarrow\quad x\bigl(4R^2 - 3x^2\bigr) = 0. ] Finally, the maximal volume: Discarding the trivial solution
The vertices of the box lie on the sphere, so each corner satisfies the equation
[ 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}}. ] It was about translating a three‑dimensional picture into
She realized that the story of Exercise 179 wasn’t just about finding a maximum volume. It was about translating a three‑dimensional picture into algebra, about the elegance of a single variable governing a whole family of shapes, and about the quiet satisfaction that comes from turning a “hard problem” into a “solved puzzle”.
First, she rewrote the volume in a friendlier form for differentiation: Finally, the maximal volume: Discarding the trivial solution