1960 AMC 12 Problem 23

Question

The radius

R

of a cylindrical box is

8

inches, the height

H

is

3

inches. The volume

V=\pi R^2H

is to be increased by the same fixed positive amount when

R

is increased by

x

inches as when

H

is increased by

x

inches. This condition is satisfied by:

no real value of

x

one integral value of

x

one rational, but not integral, value of

x

one irrational value of

x

two real values of

x

Write the volume increase once with radius

8+x

and once with height

3+x

Equate the increases and discard the nonpositive solution

Increasing the radius changes the volume by

3\pi\big((8+x)^2-8^2\big),

while increasing the height changes it by

64\pi x.

Equating these gives

48x+3x^2=64x,

so

x=0

or

x=\frac{16}{3}.

The increase must be positive, leaving one rational but nonintegral value.

Thus, the correct answer is C.

Problem 23 in other years