algebra
posted by Wen .
The horsepower (hp) that a shaft can safety transmit varies jointly with its speed (in revolutions per minute, or rpm) and the cube of the diameter. If a shaft of a certain material 2 in in diameter can transmit 36 hp at 75 rpm, what diameter must the shaft have in order to transmit 150 hp at 20 rpm?

just plug and chug:
hp = k*s*d^3
36 = k * 75 * 2^3
k = 36/(75 * 8) = 36/600
now, for the unknown d
150 = k * 20 * d^3
150 = 36/600 * 20 * d^3
150 * 600/36 * 1/20 = d^3
125 = d^3
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