Finding height of rectangular prism

idspeer

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Question: Mark has 42 identical cubes, each with 1-cm edges.He glues them together to form a rectangular solid. Ifthe perimeter of the base is 18 centimeters, find the height of the rectangular solid, in cm.

My solution: I used the equations x + y = 9 derived from 2x+2y = 18, found via perimeter requirements, x*y*h = 42, and the restriction that 42%(x*y)==0.
I made a table that listed possible values of x and y. This included:
XY
81
72
63
54

There's no need to continue the table for obvious reasons. From here I guessed and checked. The only numbers that followed the restriction 42%(x*y)==0, were x=7 and y=2. From here we know that the height is 3. (42/14 = 3)

My question:The method that I used involves guess and check. While it does solve the problem, I'd prefer a method that doesn't involve guess and check. I know that if I had another equation, I could solve the problem. I also believe there is a way to solve it with matrices. It's been a while since I've had a math class and was looking for either a solution (this isn't a hw problem for me and I already have a working solution) or direction on how to find the third equation.

(Does the third equation involve surface area?)

Thanks!

Edit: Let me know if I need to clarify any part.
 
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I'm not sure you can develop a third equation independent of the two you have (the perimeter and volume equations) since you don't really have any other information. However you can eliminate a couple of numbers from your list since both x and y also have to divide 42, i.e. not only 42%(x*y)==0 but 42%(x)==0 and 42%(y)==0. So only the pairs (2, 7) and (3, 6) need be checked.
 
I'm not sure you can develop a third equation independent of the two you have (the perimeter and volume equations) since you don't really have any other information. However you can eliminate a couple of numbers from your list since both x and y also have to divide 42, i.e. not only 42%(x*y)==0 but 42%(x)==0 and 42%(y)==0. So only the pairs (2, 7) and (3, 6) need be checked.

Yes, this also works, but it is relatively the same method (guess and check). I do appreciate the reply. Thanks!
 
Question: Mark has 42 identical cubes, each with 1-cm edges.He glues them together to form a rectangular solid. Ifthe perimeter of the base is 18 centimeters, find the height of the rectangular solid, in cm.

My solution: I used the equations x + y = 9 derived from 2x+2y = 18, found via perimeter requirements, x*y*h = 42, and the restriction that 42%(x*y)==0.
I made a table that listed possible values of x and y. This included:
XY
81
72
63
54

There's no need to continue the table for obvious reasons. From here I guessed and checked. The only numbers that followed the restriction 42%(x*y)==0, were x=7 and y=2. From here we know that the height is 3. (42/14 = 3)

My question:The method that I used involves guess and check. While it does solve the problem, I'd prefer a method that doesn't involve guess and check. I know that if I had another equation, I could solve the problem. I also believe there is a way to solve it with matrices. It's been a while since I've had a math class and was looking for either a solution (this isn't a hw problem for me and I already have a working solution) or direction on how to find the third equation.

(Does the third equation involve surface area?)

Thanks!

Edit: Let me know if I need to clarify any part.
As Ishuda pointed out you have three variables and only 2 equation so you will have to use trial and error.

If you used a 3rd equation for the surface area you would be introducing at least one new variable--the one for the surface area!
 
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You can have more than 1 solution also for all-integer cases;
example (perimeter = 50, volume = 900) :
x + y = 25 ; xyh = 900

x=15, y=10, h=6 : 15*10*6=900
x=20, y=5, h=9 : 20*5*9 = 900

Thanks for the thought. Although, is there really no other way to go about solving this question? At this point I understand there isn't a third equation.

I could try solving it with optimization calculus. :???:

Jomo said:
As Ishuda pointed out you have three variables and only 2 equation so you will have to use trial and error.

If you used a 3rd equation for the surface area you would be introducing at least one new variable--the one for the surface area!


You're right! I thought there might be a way to substitute in for the surface area, but there isn't any restriction that it is a cube (which would make that possible).
 
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