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(B) Let $R = 22 \ cm$ be the radius of the sphere and $h$ be the height of the cylinder inscribed in it. Let $r$ be the radius of the cylinder.From th

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$22\sqrt{2} \ cm$

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(B) Let $R = 22 \ cm$ be the radius of the sphere and $h$ be the height of the cylinder inscribed in it. Let $r$ be the radius of the cylinder.From the geometry of the sphere,we have $r^2 + (h/2)^2 = R^2$,which implies $r^2 = R^2 - h^2/4$.The curved surface area $A$ of the cylinder is given by $A = 2\pi rh = 2\pi h \sqrt{R^2 - h^2/4} = \pi h \sqrt{4R^2 - h^2}$.To maximize $A$,we maximize $A^2 = \pi^2 h^2 (4R^2 - h^2) = \pi^2 (4R^2h^2 - h^4)$.Let $f(h) = 4R^2h^2 - h^4$. Differentiating with respect to $h$,we get $f'(h) = 8R^2h - 4h^3$.Setting $f'(h) = 0$,we get $4h(2R^2 - h^2) = 0$. Since $h > 0$,we have $h^2 = 2R^2$,so $h = R\sqrt{2}$.Given $R = 22 \ cm$,the height $h = 22\sqrt{2} \ cm$.

If the curved surface area of a right circular cylinder inscribed in a sphere of radius $22 \ cm$ is maximum,then the height of the cylinder will be:

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