1960 AHSME Problems/Problem 23

Problem

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:

$\textbf{(A)}\ \text{no real value of x} \qquad \\ \textbf{(B)}\ \text{one integral value of x} \qquad \\ \textbf{(C)}\ \text{one rational, but not integral, value of x} \qquad \\ \textbf{(D)}\ \text{one irrational value of x}\qquad \\ \textbf{(E)}\ \text{two real values of x}$

Solution

Since increasing the height by $x$ inches should result in the same volume as increasing the radius by $x$ inches, write an equation with the two cylinders (one with height increased, one with radius increased). $\pi (8+x)^2 \cdot 3 = \pi \cdot 8^2 \cdot (3+x)$ $3(x^2+16x+64) = 64(x+3)$ $3x^2+48x+192=64x+192$ $3x^2-16x=0$ $x(3x-16)=0$ By the Zero-Product Property, $x = 0$ or $\frac{16}{3}$ . However, since $x$ must be increased, discard $x = 0$ as a possible value. Thus, the length should be increased by $\frac{16}{3}$ inches, so the answer is $\boxed{\textbf{(C)}}$ .

1960 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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1960 AHSME Problems/Problem 23

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