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Bear_B
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I have solved this problem but still have a question about it (problem and my solution posted below). What I wanted to do was express the problem as a function and use an optimization technique that would require taking the first derivative of the equation. The problem I ran into is: if using only positive integers, any function I could write out would have infinite discontinuities. Also, it does not seem like I could find the derivative of the function based on the definition of a derivative. Then I considered writing a discrete function, f(n). However, with the exception of series and sequences, I don't know how to deal with a discrete expression. I don't need a mathematical solution, but would just like to know the problem solving strategy that one would go through to optimize this problem similar to what I wanted to do.
Two adjacent faces of a rectangular box have areas 36 and 63. If all three dimensions are positive integers, find the ratio of the largest possible volume of the box to the smallest possible volume of the box.
V = l*w*h
Adjacent sides with areas 36 and 63 have a common side, which I will call h since that's how I drew it. Since all side lengths are integers, we need only consider the factors of 36 and 63.
Factors of 36 are: 18 * 2, 12 * 3, 9 * 4, and 6 * 6
Factors of 63 are: 21 * 3 and 9 * 7
The only two numbers common in both sets of factors are 3 and 9. Therefore, the largest volume box has h of either 3 or 9 and the smallest volume box has h of which ever (3 or 9) wasn't used for the largest volume box.
Using 3 for h: 3 * 12 * 21 = 756
Using 9 for h: 9 * 4 * 7 = 252
Ratio of the largest possible volume of the box to the smallest possible volume of the box:
756:252 -> 756/252 = 3/1
A simple geometric problem when approached like this, but I don't really like doing it this way because it seems boring.
[Edit: Spelling and format]
Homework Statement
Two adjacent faces of a rectangular box have areas 36 and 63. If all three dimensions are positive integers, find the ratio of the largest possible volume of the box to the smallest possible volume of the box.
Homework Equations
V = l*w*h
The Attempt at a Solution
Adjacent sides with areas 36 and 63 have a common side, which I will call h since that's how I drew it. Since all side lengths are integers, we need only consider the factors of 36 and 63.
Factors of 36 are: 18 * 2, 12 * 3, 9 * 4, and 6 * 6
Factors of 63 are: 21 * 3 and 9 * 7
The only two numbers common in both sets of factors are 3 and 9. Therefore, the largest volume box has h of either 3 or 9 and the smallest volume box has h of which ever (3 or 9) wasn't used for the largest volume box.
Using 3 for h: 3 * 12 * 21 = 756
Using 9 for h: 9 * 4 * 7 = 252
Ratio of the largest possible volume of the box to the smallest possible volume of the box:
756:252 -> 756/252 = 3/1
A simple geometric problem when approached like this, but I don't really like doing it this way because it seems boring.
[Edit: Spelling and format]
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