62 - 63 Maxima and minima: cylinder inscribed in a cone and cone inscribed in a sphere


A D V E R T I S E M E N T


Problem 62

Inscribe a circular cylinder of maximum convex surface area in a given circular cone.

 

Solution:
Cylinder inscribed in a coneBy similar triangle:
\dfrac{H - h}{d} = \dfrac{H}{D}

h = H - \dfrac{Hd}{D}

 

Convex surface area of the cylinder:
A = \pi d \, h

A = \pi d \left( H - \dfrac{Hd}{D} \right)

A = \pi H d - \dfrac{\pi H}{D}d^2

The cone is given, thus H and D are constant
\dfrac{dA}{dd} = \pi H - \dfrac{2\pi H}{D}d = 0

\pi H = \dfrac{2\pi H}{D}d

d = \frac{1}{2} D

diameter of cylinder = radius of cone            answer

 

Problem 63

Find the circular cone of maximum volume inscribed in a sphere of radius a.

 

Solution:
063-cone-inscribed-in-sphere.jpgVolume of the cone:
V = \frac{1}{3} \pi r^2 h

 

From the figure:
r^2 = a^2 – (h – a)^2
r^2 = a^2 – (h^2 – 2ah + a^2)
r^2 = 2ah – h^2

 

V = \frac{1}{3} \pi \, ( \, 2ah – h^2 \,) \, h
V = \frac{1}{3} \pi \, ( \, 2ah^2 – h^3 \,)

The sphere is given, thus radius a is constant.
\dfrac{dV}{dh} = \frac{1}{3} \pi \, ( \, 4ah – 3h^2 \,) = 0

4ah = 3h^2
h = \frac{4}{3} a

altitude of cone = 4/3 of radius of sphere            answer

 




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