Question about variation...

Question

The horsepower that a shaft can safely transmit varies jointly with its speed (in revolutions per minute) and the cube of its diameter. By what percent will the horsepower change if the speed is increased 20% and the diameter is decreased 10%. Round to the nearest tenth of a percent.

I'm sort of confused on this question because it doesn't give me any real numbers besides percentages...

So far I set up the equation as Hp=ksd^3

Hp for horsepower, s as speed and d^3 as diameter cubed.

Answer 1

Yes Indeed, Hp = k*s*d^3

20% increase means 100%+20% = 120% = 1.2

10% increase means 100% + 10% = 110% = 1.1

Replaces s = 1.2S and d=1.1D

Hp = k (1.2S)(1.1D)^3

= k * 1.5972 s*D^3

The horsepower increase by 597% as shown in bold above