A x B x C = X .-> How do I find A, B and C when X = 1.5X?

[jpardoe]

Hey there, I'm really struggling with this! I have an equation where A x B x C = X. So for example:

20 x 200 x 2 = 8,000.

I need to know the values of A, B and C when 8,000 increases. I appreciate these could be a range, so I'm happy for each one to increment by the same percentage. So for example, I'd want to reach:

22.89 x 228.94 x 2.29 = 12,000

I am unable to figure out the calculation to identify by how much A, B and C should increase in this quadratic equation. Any help would be hugely appreciated!!

 

[BigBeachBanana]

Hey there, I'm really struggling with this! I have an equation where A x B x C = X. So for example:

20 x 200 x 2 = 8,000.

I need to know the values of A, B and C when 8,000 increases. I appreciate these could be a range, so I'm happy for each one to increment by the same percentage. So for example, I'd want to reach:

22.89 x 228.94 x 2.29 = 12,000

I am unable to figure out the calculation to identify by how much A, B and C should increase in this quadratic equation. Any help would be hugely appreciated!!

1) I don't see a quadratic equation
2) If the right-hand side increases by 1.5, then the left increases by 1.5, while A, B, & C stay the same:

\(\displaystyle A\times B\times C=8,000\\ 1.5\times A\times B\times C=1.5 \times 8,000\\ 1.5\times A\times B\times C=12,000\\\)
3) Don't understand your example
4) Not sure what you're asking for ¯\_(ツ)_/¯

 

[jpardoe]

Thanks - let me try again

We know the following:

A = 20
B = 200
C = 2

Those multiplied together = 8,000

We then know that 8,000 turns to 12,000. When that happens, how do we calculate the values for A, B and C, assuming they all increase by the same percentage?

I.e. what do the new values of A, B and C become if the result is 12,000? How is that calculated?

Does that make more sense?

Thanks!

 

[Dr.Peterson]

Hey there, I'm really struggling with this! I have an equation where A x B x C = X. So for example:

20 x 200 x 2 = 8,000.

I need to know the values of A, B and C when 8,000 increases. I appreciate these could be a range, so I'm happy for each one to increment by the same percentage. So for example, I'd want to reach:

22.89 x 228.94 x 2.29 = 12,000

I am unable to figure out the calculation to identify by how much A, B and C should increase in this quadratic equation. Any help would be hugely appreciated!!

The increase could be divided in any way you like among the three factors. For instance, you could keep A and B the same and multiply C by 1.5.

If you want to increase all three by the same factor, then the cube of that factor would have to equal 1.5: [math](kA)(kB)(kC) = k^3(ABC)[/math]
So you'd want [imath]k=\sqrt[3]{1.5}=1.1447[/imath]; then the numbers would be 22.89, 228.94, and 2.29. So that must be where your numbers came from.

 

[JeffM]

I too am uncertain what you are asking.

You undoubtedly know that to solve an equation with n unknowns requires n equations.

abc = 8000 has an infinite number of solutions. Moreover, it is a cubic equation rather than quadratic.

The only thing we can say for certain about a, b, and c is that none of them is zero.

The most common way to think about your issue is difference equations.

[math]p = a + u,\ q = b + v,\ r = c + w, \text { and } pqr =12000 \implies\\ 12000 = ( a + u)qr = aqr + qru = ar(b + v) + ru(b + v) =\\ abr + avr + bru + ruv = ab(c + w) + av(c + w) + bu(c + w) + uv(c + w) =\\ abc + abw + acv + avw + bcu + buw + cuv + uvw.[/math]
Now divide that equation by 8000 = abc.

[math]\dfrac{12000}{8000} = \dfrac{abc + abw + acv + awv + bcu + buw + cuv + uvw}{abc} \implies\\ 1.5 = 1 + \dfrac{w}{c} + \dfrac{v}{b} + \dfrac{vw}{bc} + \dfrac{u}{a} + \dfrac{uw}{ac} + \dfrac{uv}{ab} + \dfrac{uvw}{abc} \implies \\ 1.5 = 1 + \dfrac{u}{a} + \dfrac{v}{b} + \dfrac{w}{c} + \dfrac{u}{a} * \dfrac{v}{b} + \dfrac{u}{a} * \dfrac{w}{c} + \dfrac{v}{b} * \dfrac{w}{c} + \dfrac{u}{a} * \dfrac{v}{b} * \dfrac{w}{c}.[/math]
Great! Now we have one cubic equation with six unknowns. An infinite number of answers. But we can simplify the problem by following Dr. Peterson’s approach. If all those fractions on the right are equal to e, we get

[math]1.5 = 1 + 3e + 3e^2 + e^3 = (1 + e)^3 \implies\\ (1 + e) = \sqrt[3]{1.5} \approx 1.145 \implies\\ e \approx 14.5\%.[/math]
This method generalizes.

 

[jpardoe]

Thank you so much for this guys! This makes perfect sense and now I see why I was struggling so much with it!

Thanks again, I really appreciate you taking the time to help. Have an awesome day ahead!