If the diameter of a right cylindrical can with circular bases is increased by \(36\%\), by what percent should the height be decreased in order to preserve the volume of the original can?

Guest Apr 16, 2022

#1**+2 **

Let the original volume = pi * r^2 * original height

Note that if the diameter is increased by 36%, then so is the radius

So

pi * r^2 * original height = pi * (1.36r) ^2 * new height

original height = (1.36)^2 * new height

original height / (1.36)^2 =new height

original height * ( 1 /1.36^2) = new height

.54 original height = new height

The original height should be reduced by ≈ (1 - .54) = .46 ≈ 46%

CPhill Apr 16, 2022

#2**0 **

Without loss of generality, let the diameter be 100, and let the height be 100.

The original volume is: \(250000 \pi\)

With the increased base, the volume of the new cube is: \(462400\pi\)

For the volume to stay the same, we have to multiply the height by \({250000 \over 462400} = {625 \over 1156}\)

This means that the percent decrease in the height is \(\approx \color{brown}\boxed{45.93 \text{%}}\)

BuilderBoi Apr 16, 2022

#3**+1 **

Let \(h\) be the original height and suppose a decrease of \(k\%\) is required.

Let \(r\) be the original radius,

Now,

\(\pi r^2 h = \pi (r(1 + 36\%))^2(h(1 - k\%))\\ (1.36)^2 (1 - k\%) = 1\\ k\% = 1 - \dfrac1{1.36^2}\\ k\% = \dfrac{531}{1156}\\ k = \dfrac{13275}{289}\)

Therefore, the height needs to be decreased by \(\dfrac{13275}{289}\%\), which is approximately \(45.9\%\).

MaxWong Apr 16, 2022