Answer to Question #148340 in Calculus for Angelo

Question #148340
Find the dimensions of the right-circular cylinder of greatest lateral surface area that can be
inscribed in a sphere of radius 6 inches.
1
Expert's answer
2020-12-08T15:07:41-0500

Let "r = 6" inches. R and h be radius and height of the cylinder.

Let R and h are related to the r by the relation,

"R = rcos\\theta, h=rsin\\theta" where "\\theta" is the angle between radius of the cylinder and radius of the sphere.


Then lateral surface area is given by, "S = 2\\pi rh = 2\\pi R^2 sin\\theta cos\\theta"

For maximum or minimum, "\\frac{dS}{d\\theta } = 0"


then, "\\frac{dS}{d\\theta } = 2\\pi R^2(cos^2\\theta-sin^2\\theta) = 0"


solving for "\\theta" , we get, "2\\pi R^2(cos^2\\theta-sin^2\\theta) = 0"

"cos^2\\theta=sin^2\\theta"

"tan\\theta = 1 \\implies \\theta = \\frac{\\pi}{4}"


Then, radius of the cylinder is

"r = Rcos\\theta = Rcos\\frac{\\pi}{4} = \\frac{6}{\\sqrt{2}} = 3\\sqrt{2} = 4.24" inches.


Height of the cylinder is,

"h = Rsin\\theta = Rsin\\frac{\\pi}{4} = \\frac{6}{\\sqrt{2}} = 3\\sqrt{2} = 4.24" inches



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