   Chapter 4.7, Problem 33E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# A right circular cylinder is inscribed in a sphere of radius r. Find the largest possible surface area of such a cylinder.

To determine

To find: The largest possible surface area of a cylinder which is inscribed in a sphere of radius r.

Explanation

Result used:

Pythagorean Theorem: If c denotes the length of the hypotenuse and a and b denote the length of the other two sides of a right angle triangle, then the Pythagorean theorem can be expressed as the Pythagorean equation: a2+b2=c2 .

Calculation:

A cylinder is inscribed in a sphere of radius r.

In Figure 1, a cylinder of height h and surface radius x, is inscribed inside a circle with radius r,

Consider the triangle from Figure 2 and apply Pythagorean Theorem as shown below.

r2=x2+(h2)2h2=r2x2h=2r2x2

The surface area of the cylinder is S(x)=2πx2+2πxh .

Substitute h=2r2x2 in S(x)=2πx2+2πxh ,

S(x)=2πx2+4πxr2x2

Differentiate S(x) with respect to x,

S(x)=4πx+4π(r22x2)r2x2

For critical values, S(x)=0 .

4πx+4π(r22x2)r2x2=0xr2x2=(r22x2)x2(r2x2)=(r22x2)2

x2r2x4=r44r2x2+4r4 (1)

Simplify the terms further, r45r2x2+5x4=0

Obtain the roots by using quadratic formula.

x2=5r2±25r445r425=5r2±25r420r410=5r2±5r410=5r2±r2510

Here, x2=5r2+r2510 does not satisfy the Equation (1).

x2=5r2r2510 is the root of Equation (1).

Hence, x=5r2r2510

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