Find the dimensions of the largest circular cone that can be inscribed in a sphere of radius R.
$V = \frac{1}{3}\pi r^2 h$
$r^2 = R^2 - (h - R)^2$
$r^2 = R^2 - (h^2 - 2hR + R^2)$
$r^2 = 2hR - h^2$
$V = \frac{1}{3}\pi (2hR - h^2)h$
$V = \frac{1}{3}\pi (2h^2R - h^3)$
$\dfrac{dV}{dh} = \frac{1}{3}\pi (4hR - 3h^2) = 0$
$4R - 3h = 0$
$h = \frac{4}{3}R$ ← height of cone
$r^2 = R^2 - (\frac{4}{3}R - R)^2$
$r^2 = R^2 – \frac{1}{9}R^2$
$r^2 = \frac{8}{9}R^2$
$r = \frac{2}{3}\sqrt{2}R$ ← radius of cone
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$V = \frac{1}{3}\pi r^2 h$
$r^2 = R^2 - (h - R)^2$
$r^2 = R^2 - (h^2 - 2hR + R^2)$
$r^2 = 2hR - h^2$
$V = \frac{1}{3}\pi (2hR - h^2)h$
$V = \frac{1}{3}\pi (2h^2R - h^3)$
$\dfrac{dV}{dh} = \frac{1}{3}\pi (4hR - 3h^2) = 0$
$4R - 3h = 0$
$h = \frac{4}{3}R$ ← height of cone
$r^2 = R^2 - (\frac{4}{3}R - R)^2$
$r^2 = R^2 – \frac{1}{9}R^2$
$r^2 = \frac{8}{9}R^2$
$r = \frac{2}{3}\sqrt{2}R$ ← radius of cone
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