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[Christoff Yurievich]

so I have two equations (cost and revenue) each with three variables. I have to find the break-even point for these equations, and I'm having a hell of a time doing it. I thought about putting the equations into a matrix, but I don't know if it's possible. I tried it and I got two answers, not three. maybe I'm just doing it wrong or something.

I tried looking in the book for something similar, but none of the problems or examples are even remotely similar to this question. *sigh* I think it's readily apparent that my calc teacher hates me. Or maybe I just hate calc. Oh well, the quarter is over in a week (final is next monday) so I'm almost finished bugging the forum for help.

EDIT: never mind that I need "halep" spelling as well...

 

[NetMapel]

Which course is this ? Micro economics ? If so, type ou the entire problem if possible...

 

[Christoff Yurievich]

It's calculus for business majors, so I suppose it can be seen as a calc-lite that leads into statistics and economics.

C(x,y,z) = 150x + 85y + 55z + 5800
R(x,y,z) = 210x + 150y + 45z

I already have P(x,y,z) (that was easy). Part c of the problem is "find the break-even point (there may be more than one but just find one of them)". The book explains that to find the break-even point you set C = R and solve. Well, with three variables that's kind of difficult. I went through google but most of that dealt with two-variables, not three.

 

[Phoenix]

I won't do it for you, but here is how you do it

C=R

1) Group all like terms together so you end up with one equation
2) Divide each side by terms so you can end up with one variable on the LHS

 

[Dilbert]

What is P(x,y,z)?

 

[Christoff Yurievich]

P(rofit) is R(evenue)-C(ost), so it ends up being 60x + 20y - 10z - 5800.

and I tried putting R = C but that just gives me P all over again. With that equation if I solve and end up with only the one variable on one side, that still leaves me with two variables on the other.

 

[NetMapel]

If it's at a equilibrium, then shouldn't profit = 0 ? If Revenue = Cost, then Profit = 0. Therefore, P = 0, collect like terms, and you've got three equations for three unknowns. Do the math !

 

[Dilbert]

Since P is not linearly independent of R and C, you still only have two independent equations for a three-variable problem. You can reduce it down to a linear equation, but that's it unless you have a third, independent equation which represents an additional constraint on the solution space.

Is it possible that the solution space is continuous? Part c of the problem is "find the break-even point (there may be more than one but just find one of them)." In that case, any point along the break-even line would be a solution...

 

[Christoff Yurievich]

I started looking for a 3D grapher program to take a look at the graphs of these equations but I can't find any that are free. I'm not surprised either.

I'll try these suggestions and see what happens. thanks.

 

[nitewulf]

since you have got an equation for profit (as you say), the break even point is when P = 0, so find that point/points.

 

[WHOAguitarninja]

Here's one way. As others have said, combine all the terms. Subtract one side from the other and you'll be left with

60x + 65y - 10z = 5800

then divide by 5800 and you'll have

(5/580)x + (13/1160)y - (1/580)z = 1

Now it becomes apparent that if you can make each term =1 when multiplied by it's coefficient, you'd be left with 1 + 1 - 1 = 1. This would satisfy the equation you created relating C and R, so it would be a solution.

All you need to realise is one solution is to set each variable equal to the reciprocal of it's constant, so you'll have

(580/5 , 1160/13 , 580) as a break even point.


EDIT - You should also check my fractions cuz I tried to do them in my head so they're probably wrong.

 

[Diablos]

What

 

[WHOAguitarninja]

What

If that was directed at me, maybe I'll try and go back and explain each step a little better.

Here's one way. As others have said, combine all the terms. Subtract one side from the other and you'll be left with

60x + 65y - 10z = 5800

What is going on here is that, as already mentioned, P= R - C So when P = 0, R=C

So that leaves us with

150x + 85y + 55z + 5800 = 210x + 150y + 45z


If you move the left side over to the right side, leaving only the constant term(5800), you should get something like

60x + 65y - 10z = 5800

then divide by 5800 and you'll have

(5/580)x + (13/1160)y - (1/580)z = 1

This is just deviding each side to get something equal to one. It's nice to set things equal to one, even if it makes your fractions ugly. You'll see why in a second...

Now it becomes apparent that if you can make each term =1 when multiplied by it's coefficient, you'd be left with 1 + 1 - 1 = 1. This would satisfy the equation you created relating C and R, so it would be a solution.

Basically what I'm saying here, is observe that if we call a, b and c the constants that are infront of x y and z respectively, we can find a solution by finding x, y, and z such that

ax = 1 by = 1 cz = 1

that would imply that

ax + by - cz = 1 This equation when you put in your fractions is just

(5/580)x + (13/1160)y - (1/580)z = 1

or if you prefer

60x + 65y - 10z = 5800

Setting it equal to one is actually simpler as the remaining steps become more intuitive.

All you need to realise is one solution is to set each variable equal to the reciprocal of it's constant, so you'll have

(580/5 , 1160/13 , 580) as a break even point.

Okay, so if ax = 1, by = 1 and cz = 1, then that implies that x=1/a y=1/b and z=1/c

if x=1/a, etc, then that means that the variable is just the reciprocal of it's coefficient. Thus....we have our answer =)

 

[retardboy]

so many numbers... im scared! although i took the class before and it does look familiar. hahaha too bad im one of those people that forgets everything right after the test.