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Container Height and Volume


Date: 08/01/97 at 16:17:09
From: Ashley Hemingway
Subject: Volume problem

How in the world do you solve this problem? 

A container's height is increased by 4 cm, and the length and width 
remain the same. If this change increased the volume by 12 percent, 
what was the original height of the container?

I would like to tell you how much Dr. Math has helped me to mature and 
grow in the field of mathematics. Thank you.

Yours truly,
Ashley Hemingway :)


Date: 08/01/97 at 18:34:39
From: Doctor Mike
Subject: Re: Volume problem

Thanks for the kind words about our help in the past.  I'll try to
keep up the good work.  

From the way the problem is stated, I assume the container is box-
shaped, so the volume is the product of length and width and height. 

Let the three original values be L, W, and H.  Then the original 
volume is L*W*H.  The new volume is L*W*(H+4), but it also equals 
the original volume times 1.12 since it has 12 percent more volume.  
So you get the equation  L*W*(H+4) = L*W*H*1.12  which you can solve 
for H.  

First divide both sides of the equation by L*W to get an equation with 
H as the only variable.   
   
Note that L*W is the area "B" of the base of the box.  The problem 
would be worked in essentially the same way if it were stated that 
"A container's height is increased by 4 cm, and the base remains
exactly the same."  Then the equation is B*(H+4) = B*H*1.12 and you
proceed in the same way.  You don't really need to know anything
about the shape of the bases.  They could both be squares, or both
be circles, or both be equilateral triangles, or whatever. 
    
What I did in the previous paragraph is to make this problem just a
little more general, but keep the idea of how to solve it the same. 
Your increased maturity will help you to "look at the bigger picture"
like this. The situation could be made even more general by changing
the problem from "the base remains exactly the same" to "the area of
the base remains exactly the same".  The first container could be a
rectangular-based box and the second container a circular-based can,
but if both of them have the same AREA then we can still use the
equation B*(H+4) = B*H*1.12 where B is that common area.  
   
I hope this helps.  If you have more questions, write back to us.

-Doctor Mike,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry

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