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The volume of a box is 2208 cm3 The surface area is 480 cm2. What is the height of the box in centimeters?

 Mar 17, 2015

Best Answer 

 #2
avatar+118608 
+13

Hi Chris,

I have  looked through all of your working and it all looks good to me.     (๑‵●‿●‵๑)

 

It is good that you went to all tihs effort.

 

I would just have said

With three unknowns you need three equations.  Since there are only two equations you cannot find a single numerical answer.  In this case you have found the answer - the answer does not exist - kind of funny don't you think?

 Mar 18, 2015
 #1
avatar+128406 
+13

There is no answer to this......here's why....

The surface area is given by

Sa  = 480 =  2(lw + wh + lh)

240 = w( l + h) + lh

w = [240 - lh] / (l + h)

 

And the volume is given by

V = 2208 = lwh

2208/ [lh] = w

 

And setting the "w's"  equal, we have

2208/[lh] = [240- lh] / (l + h)    ..... cross-multiply

2208(l + h) = [240 - lh] lh

2208(l + h) = 240lh - (lh)^2     set to 0

(lh)^2 + 2208(l + h) - 240lh = 0

Now.....letting x = l and h = y.....look at the graph of

(xy)^2 + 2208(x + y) - 240xy = 0

https://www.desmos.com/calculator/eocxww2vmh

Notice that l and h both have to be greater than 0 because they are both lengths. Thus, x and y have to be greater than 0.

But notice.......if x and y were greater than 0, at least part of the graph would have to exist in the first quadrant....but it doesn't

Thus, there are no solutions......

 

P.S.  .....would another mathematician check my logic here???......I believe this is correct, but I'm not sure.....!!!!

  

 Mar 17, 2015
 #2
avatar+118608 
+13
Best Answer

Hi Chris,

I have  looked through all of your working and it all looks good to me.     (๑‵●‿●‵๑)

 

It is good that you went to all tihs effort.

 

I would just have said

With three unknowns you need three equations.  Since there are only two equations you cannot find a single numerical answer.  In this case you have found the answer - the answer does not exist - kind of funny don't you think?

Melody Mar 18, 2015
 #3
avatar+4 
+5

I was the original questioner. (Do i say that right?)

 

I want to really thank you CPhill for your support, and your effort you put into that.

 

 

Thank you very much.

 

~Damian

 Mar 20, 2015
 #4
avatar+128406 
0

Thanks, Damian......that was a good problem, too.....a little bit of a challenge.....!!!!

 

  

 Mar 20, 2015

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