Math help needed - Calculus

[rust0r]

Hey guys, hoping someone might be able to give me a hand with this problem from my university business class. If someone is able to provide the answer/workthrough and a little explanation that would be much appreciated. Thanks!


Consider:

q1 = 95 - p1 + 0.7p2
q2 = 188 + 0.03p1 - 5p2

where q1 and q2 are two substitutable products with p1 and p2 as their prices respectively

c1 = 100 + 90q1
c2 = 50 + 250q2

c1 and c2 is the cost of producing each

Find the optimal solution



We started it in class and ended up with:

Profit = TR - TC (obviously)
= p1q1 + p2q2 - c1 - c2

 

[jerrad]

God I hate math.
GL man, glad it's you and not me. lol

 

[txyaloo]

You find the quantities produced with the second set of formulas. Use that to find the price with the first set of formulas. After you have price, you can find cost and then find profit.

To find the optimal mix, you'll need to compute the marginal cost and marginal revenue. MR=MC is the optimal mix.

 

[mach1won]

you lost me at calculus....
good luck though man!! haha

 

[thomas91169]

**** that I quit.

 

[rust0r]

You find the quantities produced with the second set of formulas. Use that to find the price with the first set of formulas. After you have price, you can find cost and then find profit.

To find the optimal mix, you'll need to compute the marginal cost and marginal revenue. MR=MC is the optimal mix.

Hmm, I guess I'm a little more lost on the problem than I let on lol. Thanks for that though

The prof gave this as an extra credit question as it isn't something we normally do, so I'm a little unsure of how to proceed based on you explanation

 

[UFGatorGuy20]

Enter all your formulas into Profit = p1q1 + p2q2 - c1 - c2. When you do that and multiply everything out, you will have a polynomial in terms of p1 squared, p2 squared, p1p2, q1, and q2. Substitute q1 and q2 back in so that you have an equation completely in terms of p1 and p2. This long profit formula is maximized when your derivative equals zero. Problem is, you have two variables... p1 and p2.

Take partial derivative with respect to p1 (set equal to zero). Take partial derivative with respect to p2 (set equal to zero). Solve for p1 and p2.

I think that should work...

 

[rust0r]

Thanks for the help, can anyone actually provide me the answer? I'm stumbling my way through this, not sure I'm doing it right :/

 

[txyaloo]

Thanks for the help, can anyone actually provide me the answer? I'm stumbling my way through this, not sure I'm doing it right :/

Nope. Profs have office hours for a reason. oke:

 

[lobra97]

just put TRUE...

 

[Coiled03]

Enter all your formulas into Profit = p1q1 + p2q2 - c1 - c2. When you do that and multiply everything out, you will have a polynomial in terms of p1 squared, p2 squared, p1p2, q1, and q2. Substitute q1 and q2 back in so that you have an equation completely in terms of p1 and p2. This long profit formula is maximized when your derivative equals zero. Problem is, you have two variables... p1 and p2.

Take partial derivative with respect to p1 (set equal to zero). Take partial derivative with respect to p2 (set equal to zero). Solve for p1 and p2.

I think that should work...

This.

/thread

 

[rust0r]

Thanks for the help from everyone, I think I managed to get some of the marks, if not all.

Txyaloo, keeping us honest haha :beer:

 

[Camaro_94]

Numbers and letters shouldnt be mixed with eachother. You're only asking for trouble when that happens. Rule of thumb, keep them separated, and you're golden. I used to hate it when we used to have to "find x" in high school! Do I look like a pirate? Because the only people who needs to "find x" are pirates.

just put TRUE...

+1!

 

[rust0r]

. I used to hate it when we used to have to "find x" in high school! Do I look like a pirate? Because the only people who needs to "find x" are pirates.

Haha, good one

"Find X"

"ARRRR MATEEE"

 

[Camaro_94]

Haha, good one

"Find X"

"ARRRR MATEEE"

Exactly! :lol1:

 

[txyaloo]

Numbers and letters shouldnt be mixed with eachother. You're only asking for trouble when that happens. Rule of thumb, keep them separated, and you're golden. I used to hate it when we used to have to "find x" in high school! Do I look like a pirate? Because the only people who needs to "find x" are pirates.



+1!

:lol: If only that were true in real life. Too bad some people actually have to find X on a daily basis.

 

[TwinTurbo4vGT]

:lol: If only that were true in real life. Too bad some people actually have to find X on a daily basis.

 

[stangin99]

:lol: If only that were true in real life. Too bad some people actually have to find X on a daily basis.

X is easy. Z gets tricky.

 

[truebluedevil02]

Enter all your formulas into Profit = p1q1 + p2q2 - c1 - c2. When you do that and multiply everything out, you will have a polynomial in terms of p1 squared, p2 squared, p1p2, q1, and q2. Substitute q1 and q2 back in so that you have an equation completely in terms of p1 and p2. This long profit formula is maximized when your derivative equals zero. Problem is, you have two variables... p1 and p2.

Take partial derivative with respect to p1 (set equal to zero). Take partial derivative with respect to p2 (set equal to zero). Solve for p1 and p2.

I think that should work...

I concur...with whatever you just said. Honestly in HS I had to cheat my ass off to pass Geometry with a C-, lol. I LOVE physics and astronomy but the math you get into in the more advanced stages of it just makes me want to shoot myself, lol.

 

[thomas91169]

Calculus, U+Me=Us?