Discarding the trivial solution (x = 0) (which gave zero volume), she solved

[ 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}}. ]

Simplifying gave

Setting the numerator to zero (the denominator never vanished inside the feasible interval) produced

First, she rewrote the volume in a friendlier form for differentiation:

[ \left(\frac{x}{2}\right)^2+\left(\frac{x}{2}\right)^2+\left(\frac{y}{2}\right)^2 = R^2 . ]

[ 4xR^2 - 3x^3 = 0 \quad\Longrightarrow\quad x\bigl(4R^2 - 3x^2\bigr) = 0. ]