Solucionario Calculo Una Variable Thomas Finney Edicion 9 179 -

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 . ]

Using the product rule and the chain rule, she obtained

Setting the numerator to zero (the denominator never vanished inside the feasible interval) produced First, she rewrote the volume in a friendlier

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.

When she stood, the room fell silent. She described the geometry, the substitution of , the elegant reduction to a single‑variable function, and the calculus steps that led to the cube. She finished with the final expression (\displaystyle V_{\max}= \frac{8R^3}{3\sqrt{3}}) and a quick sketch of the inscribed cube inside the sphere.

[ 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}} . ] When she stood, the room fell silent

Plugging this back into the expression for :

A ripple of impressed murmurs ran through the class. The professor nodded, his eyes twinkling. “Excellent,” he said. “You’ve illustrated perfectly how a multivariable problem can sometimes be reduced to one variable, and how the critical point tells us the shape of the optimal object. Well done, Maya.”

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. [ V(x) = x^2 \cdot y = x^2

[ 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}}}. ]

Now the volume of the box was simply