# math nerds

kuzi16
Posts:

**14,633**✭✭✭✭
i have a problem for you.

i am looking for a generic equation to solve this problem:

you have two different sized boxes. you wish to stack them next to each other and have the tops of the box align and be level.

there is one catch...

one of the stacks is on a platform.

x = height of the fist box

y = height of the other box

a = height of the platform

H = height of the finished stacked boxes

for this problem lets assume that the boxes of height "y" are on the platform and that you have an infinite number of both boxes. the platform is immobile and neither of the boxes can change in size.

if there was no platform then the problem would be simple: xy=h (7 x 4 = 28 (for instance))

EDIT: im looking for the equation that will give you H, not the equation that will give you the number of each boxes needed.

i am looking for a generic equation to solve this problem:

you have two different sized boxes. you wish to stack them next to each other and have the tops of the box align and be level.

there is one catch...

one of the stacks is on a platform.

x = height of the fist box

y = height of the other box

a = height of the platform

H = height of the finished stacked boxes

for this problem lets assume that the boxes of height "y" are on the platform and that you have an infinite number of both boxes. the platform is immobile and neither of the boxes can change in size.

if there was no platform then the problem would be simple: xy=h (7 x 4 = 28 (for instance))

EDIT: im looking for the equation that will give you H, not the equation that will give you the number of each boxes needed.

0

## Comments

4,517✭For example, if the platform were 6" high:

H = 7(4+(6/7)) = 34

Or that's my guess, anyway.

412✭7,349✭201✭977H = x(y+a)

The first box is going to always = H but the y has the addition of the platform factored in.

2,555✭Thats a good one. Is this real or hypothetical?

14,633✭✭✭✭where "c" and "d" are the number of boxes needed from each? or am i now throwing in new variables that will throw everything off?

if it were h = x(y+a) then you would be adding the height of the platform every for every box "y"

14,633✭✭✭✭2,555✭977Let k = constant of box 1, J = constant of box 2, a = height of platform and x be the variable.

k(j+a) + S{(kx)(jx)

The first part of the equation is to get both box k and box j + platform a to a level surface, then the second part is the summation of k and j to any multiple of x. What this means is that this equation is good for finding any height where both boxes are level.

Here's an example using numbers:

k = 5, j = 4, a = 3

5(4+3) + S (5 x 2)(4 x 2)

35+80 = 115 inches.

Checking the answer: 115/5 = 23, 115 - 3(To account for platform height) = 112/4 = 28

2,555✭4,517✭Your formula works, I think, but you'd have to know both c & d, in addition to x, y, and a, to find out the H you were looking for. If my formula works -- and I

thinkit does, but I was never all that great at math -- you only need to know x, y, and a, the heights of the 3 objects.2,560✭✭BOX.

5,876✭✭✭✭✭14,633✭✭✭✭on that note, can you explain to me why yours works and mine wouldnt? (i have to admit, i didnt test my formula)

977The equation works because it's basically doing what your equation does but has that first part in there to compensate for the platform. That way you'll start off with a fresh, level surface before finding other heights that satisfy your requirements of being level.

Think of it as making a building, you pour a good solid foundation so that you can get as level as possible and then build up from there.

7,349✭4,473✭✭✭9,403✭