Solucionario Calculo Una Variable Thomas Finney Edicion 9 179 [ EXCLUSIVE · 2024 ]

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.

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.”

[ 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}}}. ]

A pleasant symmetry emerged: the height and the side of the base were equal! The optimal box turned out to be a whose edge length was (\frac{2R}{\sqrt{3}}). After the class, several classmates gathered around Maya,

Maya wrote the result in bold, underlined it, and added a small smiley face next to it—her personal signature of triumph. The next morning, the professor walked into the seminar room, a stack of papers in his hand. He asked the class to volunteer a solution for Exercise 179. Maya’s hand rose, heart thudding like a metronome.

[ V'(x) = 4x\sqrt{R^2 - \tfrac{x^2}{2}} - \frac{x^3}{\sqrt{R^2 - \tfrac{x^2}{2}}}. ]

Maya solved for in terms of x :

Using the product rule and the chain rule, she obtained

[ \frac{x^2}{2} + \frac{y^2}{4} = R^2. ]

[ 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 closed the library’s heavy door and stepped

[ V(x) = 2x^2 \bigl(R^2 - \tfrac{x^2}{2}\bigr)^{1/2}. ]

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.

On the central table lay a battered copy of Thomas’ Calculus, 9th edition , its corners softened by years of eager thumbs. A thin, yellowed sheet was tucked between pages 178 and 180, its header scrawled in a hurried hand: . Maya’s professor had hinted that the problem was a “real gem” and that the solution would be discussed the next week—if anyone could actually work it out. “Excellent,” he said

Simplifying gave

Factoring out the common denominator gave