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?
+128474 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%
+2666 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 \(\approx \color{brown}\boxed{45.93 \text{%}}\)
+9519 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 \(45.9\%\).
