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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?

 Apr 16, 2022
 #1
avatar+122385 
+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%

 

cool cool cool

 Apr 16, 2022
edited by CPhill  Apr 16, 2022
 #2
avatar+1351 
+1

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 \(\approx \color{brown}\boxed{45.93 \text{%}}\)

 Apr 16, 2022
 #4
avatar+1351 
+2

Ayoo... That's my 400th answer

BuilderBoi  Apr 16, 2022
 #5
avatar+9277 
+1

Nice!

MaxWong  Apr 16, 2022
 #6
avatar+122385 
+1

Whoopee !!!

 

Also.....good to see you on here, Max  !!!!!

 

 

 

cool cool cool

CPhill  Apr 16, 2022
 #7
avatar+9277 
+1

Nice to see you too, CPhill :)

It has been a long time since I was last active on this site. It feels great to come back to this site and see that it is still a very active forum.

MaxWong  Apr 16, 2022
 #8
avatar+122385 
+1

I was away about nine months, too....it was a little slow when I first came  back, but it appears to have "revived" some !!!!

 

 

cool cool cool

CPhill  Apr 16, 2022
 #3
avatar+9277 
+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 \(45.9\%\).

 Apr 16, 2022

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