Math Question For Clive. (1 Viewer)

SoCal

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Hey Clive, I know you love your Maths so I have a question for you and I am wondering if you can help me out with it. We are given this equation:

Profit(2) = (P2 - 3.94)(49.52 - 5.48P2 + 1.40P1)

and we have to differentiate Profits with respect to P2 (treating P1 as a constant) and then maximise it by setting the expression equal to 0 and solving the resulting equation for P1. The final answer should be P2 = 6.49 + .1277P1.

Maths used to be my best subject but I guess this is what happens when you don't do it for over a year, coupled with the fact that our Lecturer teaches us nothing and expects us to come up with the method ourselves:mad:. Anyway, any help would be appreciated. Thanks dude:).
 

jumb

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Merethrond said:
Hey Clive, I know you love your Maths so I have a question for you and I am wondering if you can help me out with it. We are given this equation:

Profit(2) = (P2 - 3.94)(49.52 - 5.48P2 + 1.40P1)

and we have to differentiate Profits with respect to P2 (treating P1 as a constant) and then maximise it by setting the expression equal to 0 and solving the resulting equation for P1. The final answer should be P2 = 6.49 + .1277P1.

Maths used to be my best subject but I guess this is what happens when you don't do it for over a year, coupled with the fact that our Lecturer teaches us nothing and expects us to come up with the method ourselves:mad:. Anyway, any help would be appreciated. Thanks dude:).
I got:

Profit(2) = (P2 - 3.94)(49.52 - 5.48P2 + 1.40P1)
let
a = 3.94
b = 49.52
c = 5.48
d = 1.40

Profit(2) = (P2 - a)(b - cP2 + dP1)

(looks neater now)

= bP2 - c(P2)^2 + dP2P1 -ab +acP2 + adP1
d(Profit(2))/d(P2) = b - 2cP2 + dP1 + ac = 0
2cP2 = dP1 +b + ac
P2= (d/2c)P1 + (b + ac)/2c

= 0.1277P1 + 6.49 (as required)

(after about 10 edits)
 
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SoCal

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jumb said:
I got:

Profit(2) = (P2 - 3.94)(49.52 - 5.48P2 + 1.40P1)
let
a = 3.94
b = 49.52
c = 5.48
d = 1.40

Profit(2) = (P2 - a)(b - cP2 + dP1)

(looks neater now)

= bP2 - c(P2)^2 + dP2P1 -ab +acP2 + adP1
d(Profit(2))/d(P2) = b - 2cP2 + dP1 + ac = 0
2cP2 = dP1 +b + ac
P2= (d/2c)P1 + (b + ac)/2c

= 0.1277P1 + 6.49 (as required)

(after about 10 edits)
Thanks very much for your help jumb:). So I have some questions now. Firstly, is it necessary to substitute the numbers or can you get the same solution using the numbers? I think you were just using letters to make it clearer? Anyway, I have one other question:

jumb said:
= bP2 - c(P2)^2 + dP2P1 -ab +acP2 + adP1
d(Profit(2))/d(P2) = b - 2cP2 + dP1 + ac = 0
When you differentiated every term with P1 in it do you just treat P1 as some other number because it is constant? I think you do, I am just making sure:).
 

baker182

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Oh yeah assignment 1, lucky i had the answers to that one :p

By the way, Have you done much study for Business Finance or Money and Banking?
 

SoCal

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This isn't Assignment 1 content:p!

I have done absolutely no study for Business Finance (I have only read the first two weeks readings and have not done any questions whatsoever apart from the one you asked about):(. I have read all the lecture notes for Money and Banking, which is all I can do because I didn't photocopy the other supplementary materials with the diagrams. I am going to get a copy of the graphs that show what factors affect the supply and demand for Bonds before the mid-semester exam though because I had no idea about them in the class test:p. I thought you were dropping Money and Banking thought:)?
 

baker182

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Merethrond said:
This isn't Assignment 1 content:p!

I have done absolutely no study for Business Finance (I have only read the first two weeks readings and have not done any questions whatsoever apart from the one you asked about):(. I have read all the lecture notes for Money and Banking, which is all I can do because I didn't photocopy the other supplementary materials with the diagrams. I am going to get a copy of the graphs that show what factors affect the supply and demand for Bonds before the mid-semester exam though because I had no idea about them in the class test:p. I thought you were dropping Money and Banking thought:)?
I did drop Money and Banking, just wondering what study you have done hehee. By the way, I have all the textbook chapters photocopied, if you like I could send you the diagrams you want?
 

SoCal

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OK, so how much study have you done for your subjects? Yeah dude, if you could just post up those diagrams that would be awesome. Thanks:).
 

baker182

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Merethrond said:
OK, so how much study have you done for your subjects? Yeah dude, if you could just post up those diagrams that would be awesome. Thanks:).
Business Finance, I'am about to start week 4. Coporate Accounting just finshed week 2.

I will scan the diagrams tommorrow, is that cool with you?
 

jumb

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Merethrond said:
Thanks very much for your help jumb:). So I have some questions now. Firstly, is it necessary to substitute the numbers or can you get the same solution using the numbers?
No, I just think it's easier using letters as apposed to numbers with decimals.

Merethrond said:
When you differentiated every term with P1 in it do you just treat P1 as some other number because it is constant? I think you do, I am just making sure:).
Yeah, cause like you said, it's a constant. So, any term that doesn't have any P2's will be automatically be timesed by 0.
 
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SoCal

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jumb said:
No, I just think it's easier using letters as apposed to numbers with decimals.
Yeah, I thought so, I was just making sure:).

jumb said:
Yeah, cause like you said, it's a constant. So, any term that doesn't have any P2's will be automatically be timesed by 0.
Ditto. Thanks again:).
 

evilc

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It's already been done but..anyway

Easy bit of partial differentiation here :)

call Profit(2) F(x,y)
call P2 x
call P1 y

Profit(2) = (P2 - 3.94)(49.52 - 5.48P2 + 1.40P1)

F(x,y) = 49.52x - 5.48x^2 + 1.40xy + 21.5912x - 5.488y - 195.1088

Partially differentiating the expression with respect to x (we only need to find out how the expression changes with respect to x, as y is held constant)

∂F/∂x = 49.52 - 10.96x + 1.4y + 21.5912

solve ∂F/∂x = 0 to find maximum

49.52 - 10.96x + 1.4y + 21.5912 = 0

10.96x = 71.1112 + 1.4y

x = (71.1112 + 1.4y)/(10.96)

x = 6.488 + 0.1277y :)
 
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mrbassman

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evilc said:
It's already been done but..anyway

Easy bit of partial differentiation here :)

call Profit(2) F(x,y)
call P2 x
call P1 y

Profit(2) = (P2 - 3.94)(49.52 - 5.48P2 + 1.40P1)

F(x,y) = 49.52x - 5.48x^2 + 1.40xy + 21.5912x - 5.488y - 195.1088

Partially differentiating the expression with respect to x (we only need to find out how the expression changes with respect to x, as y is held constant)

∂F/∂x = 49.52 - 10.96x + 1.4y + 21.5912

solve ∂F/∂x = 0 to find maximum

49.52 - 10.96x + 1.4y + 21.5912 = 0

10.96x = 71.1112 + 1.4y

x = (71.1112 + 1.4y)/(10.96)

x = 6.488 + 0.1277y :)
Thats how i'd have done it however i wouldnt have went to the trouble of the delta
 

SoCal

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Yeah, it is partial differentiation but the lecturer didn't really explain what that was. So is it just differentiating one variable? Also, you know how you differentiate the equation and put it equal to zero to find the maximum, doesn't that also find the minimum and you have to test to find if it is a maximum or minimum:)?
 

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