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A cylindrical chocobar has its radius $r$ unit and height $h$ unit. If we wish to increase the volume by same amount either by increasing its radius alone or its height alone by the same number of units, then how many units do we have to increase the radius or height?

a) $\dfrac{r^2+2r}{h}$

b) $\dfrac{r^2-2rh}{h}$

c) $\dfrac{2r^2-rh}{h^2}$

d) $\dfrac{\pi r^2}{2h}$

I tried to solve this in this way. $$\pi(r+x)^2h=\pi r^2(h+x)$$ $$x^2h+2rxh=r^2x$$ and then $x=\dfrac{r^2-2rh}{h}$. But the correct answer is given as option c.

How to approach towards the solution? Any help is appreciated that directs me to the solution.

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3  
+1 for showing your work! Your solution looks right to me; perhaps there's an error in the book regarding the answer, or perhaps the problem is stated incorrectly. –  Zev Chonoles Mar 5 '13 at 11:05
    
@Zev Chonoles Thanks.... but i want to confirm that my work is not erroneous. –  cdummy Mar 5 '13 at 11:09
    
Getting answer as : $ (r^2 - 2rh)/h $ . The option given might have been misprinted. –  lsp Mar 5 '13 at 11:09
    
You're good, cdummy! My answer confirms yours...Nice job with showing your work; I second Zev's +1 –  amWhy Mar 5 '13 at 11:11
    
@lsp sry now i changed the options now check it..pls –  cdummy Mar 5 '13 at 11:12
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1 Answer 1

up vote 2 down vote accepted

Change in $r \to r+x$ vs. change in $h \to h+x \implies$ volume equal in each case, so yes, and knowing the volume of a cylinder is given by $\pi r^2 h$, we have:

$$\pi(r+x)^2h=\pi r^2(h+x)$$

$$\iff (r+x)^2h = r^2(h+x) $$ $$\iff r^2h + 2xrh + x^2h = r^2 h + r^2 x$$

$$\iff x^2h+2rxh=r^2x $$ $$\iff xh+2rh = r^2 $$ $$\iff xh = r^2 - 2rh $$ $$ \iff x=\dfrac{r^2-2rh}{h}$$

So it looks like a match!

$$(b)\quad x = \frac {r^2 - 2rh}{h}$$

Looks like a match with your answer, and mine! Well done!

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yes thanks for your time. –  cdummy Mar 5 '13 at 11:17
    
You're welcome: you did very well with this. Trust yourself...you were thorough, and careful, and showed your work. Sometimes we all second guess ourselves ;-) –  amWhy Mar 5 '13 at 11:20
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