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Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085

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BuyFindarrow_forward

Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085
Textbook Problem

In calculus, it can be shown that the largest possible volume for the inscribed right circular cylinder in Exercise 26 occurs when its altitude has a length equal to the diameter length of the circular base. Find the length of the radius and the altitude of the cylinder of greatest volume if the radius length of the sphere is 6 in.

To determine

To find:

The length of the radius and altitude of the inscribed cylinder of greatest volume.

Explanation

Approach:

The surface area of a sphere S=4πr2, where r is the radius of the sphere.

Volume of a sphere V=43πr3.

For right circular cylinder of base radius (r), and altitude (h),

Total surface area of cylinder=2πr(h+r)

Curved or lateral surface area=2πrh.

Volume of cylinder=πr2h.

Calculation:

Consider a right circular cylinder is inscribed within a sphere where,

Altitude (h) = Diameter (d) = 2 · Radius of cylinder = 2(r)

Therefore h = 2r

Radius of sphere= (R) = 6 in.

The diameter of the circular bases of a right circular cylinder along with the altitude of the cylinder form a square, when the diameter and the altitude are equal in length.

When a square is circumscribed by a circle, the length of diagonal of the square is equal to the diameter of the circle. We therefore use Pythagorean theorem to find the sides of the square.

The square of the hypotenuse is equal to the sum of squares of the other two sides:

hypotenuse2=side2+side2

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