Solucionario Calculo Una Variable Thomas Finney Edicion 9 179 [100% SIMPLE]

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”.

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.

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

[ V(x) = 2x^2 \bigl(R^2 - \tfrac{x^2}{2}\bigr)^{1/2}. ] She realized that the story of Exercise 179

which simplified to

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

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. The next morning, the professor walked into the

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

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.

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

Plugging this back into the expression for :

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}}).