The vertices of the box lie on the sphere, so each corner satisfies the equation
As she walked home, she imagined the inscribed cube—edges perfectly aligned, each corner just touching the sphere—sitting like a gem inside a glass sphere, a concrete reminder that sometimes, the most beautiful solutions are the simplest, and that every calculus problem hides a story waiting to be told.
Simplifying gave
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.”
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 vertices of the box lie on the
Maya had been wrestling with the problem all semester. It was the sort of question that seemed simple at first glance, then revealed hidden layers like an onion. The statement asked her to , using only one variable. In other words, the box’s height and the side of its base were tied together by the geometry of the sphere, and the challenge was to express the volume in terms of a single unknown, then locate its critical point.
[ V'(x) = 4x\bigl(R^2 - \tfrac{x^2}{2}\bigr)^{1/2} + 2x^2\left(\tfrac{1}{2}\right)\bigl(R^2 - \tfrac{x^2}{2}\bigr)^{-1/2}(-x) ] “Excellent,” he said
Maya solved for in terms of x :
