Simple Calculus I question

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wiredspider

Diamond Member
Jun 3, 2001
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So this question is fairly simple, yet I don't get what they say is the correct answer and it is driving me a little nuts. I got a final in this class tomorrow morning that I'm sure to fail, but I guess I might as well try...

Anyhow here is the question, let me know what you get as the answer and I'll post what they say is right. I get a different answer than what they say and think I'm right, but I suppose I'm just missing something simple...

The demand function for a certain product is p=80-.5x and the cost function is C=100+30x. What price will yield the maximum profit? (x is the number of units produced)
(a) $68
(b) $24
(c) $30
(d) $55
(e) $50
 
A

Ausm

Lifer
Oct 9, 1999
25,215
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I used to do three to four 3 foot bongs before my calc class but that was 20 years ago ;)


Ausm
 
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shuan24

Platinum Member
Jul 17, 2003
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i think the problem is that there are two unknown variables: the number of units produced and the price you are selling at. Would this be the correct equation: profit = (demand * sell price) - cost ?

Makes sense to me, but I don't see where the calc is involved.
 
Platypus

Platypus

Lifer
Apr 26, 2001
31,053
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Originally posted by: shuan24
i think the problem is that there are two unknown variables: the number of units produced and the price you are selling at. Would this be the correct equation: profit = (demand * sell price) - cost ?

Makes sense to me, but I don't see where the calc is involved.
Click to expand...

From what I remember, I think you need to derive something
 
D

dighn

Lifer
Aug 12, 2001
22,820
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Originally posted by: CorporateRecreation
Originally posted by: shuan24
i think the problem is that there are two unknown variables: the number of units produced and the price you are selling at. Would this be the correct equation: profit = (demand * sell price) - cost ?

Makes sense to me, but I don't see where the calc is involved.
Click to expand...

From what I remember, I think you need to derive something
Click to expand...


with the profit function you derive with respect to x, find where derivative is 0. check values at those pints and at boundaries and find x with best profit

but what is the profit function? that's the ah heck
 
W

wiredspider

Diamond Member
Jun 3, 2001
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It's not A, I didn't get A and they don't say its A. More I think about the problem more confused I get :(. And what information given is all that is given with the problem.
 
James3shin

James3shin

Diamond Member
Apr 5, 2004
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well im going to say that you take demand function minus the cost function, take derivative, set equal to zero, find x-value, create a sign chart to see if said x-value is a max or min.....tada?
 
S

shuan24

Platinum Member
Jul 17, 2003
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the derivative of a constant * x is the constant. How do you set a constant to zero?
 
D

dighn

Lifer
Aug 12, 2001
22,820
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Originally posted by: shuan24
the derivative of a constant * x is the constant. How do you set a constant to zero?
Click to expand...

then it's never zero. but there are still the end points (+-inf). so in the case of a line the max/min are +- inf which makes sense
 
James3shin

James3shin

Diamond Member
Apr 5, 2004
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hmm, i swore i saw a exponent for those equations, my bad....but there are end points like dighn said
 
S

shuan24

Platinum Member
Jul 17, 2003
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i will agree with the pizza guy....you're definitely missing something.

edit/ care to share?
 
DrPizza

DrPizza

Administrator Elite Member Goat Whisperer
Mar 5, 2001
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edit to my edit. Something is missing.

Suppose you make 1 product, and you sell it for 1 bazillion dollars.

Huge profit! And, nothing says you can't do that. Cost is the cost of making 1 product. Demand is enough to sell it.

Oddly, though, I would have thought demand would be a function of the selling price, rather than a function of the number made...
 
S

shuan24

Platinum Member
Jul 17, 2003
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makes sense and all, but I believe there is a probelm with the equation Revenue = demand * x (where x is units made).

the fallacy is that you will sell every unit that you make.

edit: oh I see. I guess if you replace the word "demand" with "sell price" then it makes sense.
 
DrPizza

DrPizza

Administrator Elite Member Goat Whisperer
Mar 5, 2001
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p=80-.5x
p is the price, x is the demand.

logically, most of us would rather think of the equation as
x (demand in units) = 160 - 2p (the selling price)
I was thinking that d stood for demand. I don't know why I thought that way.
 
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unsped

Platinum Member
Mar 18, 2000
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the problem is this thread is worded very poorly. this isnt a calculus question so much as its an economic class question. the reason people are finding it difficult is because there are other equations that you are not including, equations you would know if you went to class ;)

im assuming you use both equations to find a condition that is known before hand as its a basic profit maximization question... depending on the type of firm, the profit max condition changes.

for instance a business in a free market situation generally has profit max at

marginal revenue = marginal cost

with the price equation you can easily figure out the revenue equation

marginal revenue is the derivative of revenue, marginal cost is the derivative of cost.

a producer will produce up to a point where the amount of revenue they get for an additional item is equal to thier costs incurred with selling that one item.

since revenue and cost are dependant on quanities, there are derivatives involved.

http://www.digitaleconomist.com/profit.html
 
S

squeeg22

Senior member
Feb 28, 2001
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Originally posted by: unsped
the problem is this thread is worded very poorly. this isnt a calculus question so much as its an economic class question. the reason people are finding it difficult is because there are other equations that you are not including, equations you would know if you went to class ;)

im assuming you use both equations to find a condition that is known before hand as its a basic profit maximization question... depending on the type of firm, the profit max condition changes.

for instance a business in a free market situation generally has profit max at

marginal revenue = marginal cost

with the price equation you can easily figure out the revenue equation

marginal revenue is the derivative of revenue, marginal cost is the derivative of cost.

a producer will produce up to a point where the amount of revenue they get for an additional item is equal to thier costs incurred with selling that one item.

since revenue and cost are dependant on quanities, there are derivatives involved.

http://www.digitaleconomist.com/profit.html

MR=P=MC
MC = dC = 30
80-.5x=30
x=100(units)

P=80-.5(100)=30



unless im mistaken, the profit max number of units is 100, plug that back into the price equation and the answer is C 30
Click to expand...


Actually, isn't the answer $50, or E?

By using the equations DrPizza found:

R=80x-.5x^2
C=100+30x

P=R-C

P=80x-.5x^2-(100+30x)
P=-.5x^2+50x-100

P'=-x+50

therefore, x=50 is the only answer
 
U

unsped

Platinum Member
Mar 18, 2000
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woops here is the right answer i forgot to turn P into TR ;)

(d) 55
when x=50

TR = P*Q
TR = 80x-.5x^2
MR = 80-x

C = 100+30x
MC = 30

set MR = MC

80-x = 30
x = 50 (quanitity)

price = 80-25x = 55
 
J

Jmman

Diamond Member
Dec 17, 1999
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Originally posted by: unsped
woops here is the right answer i forgot to turn P into TR ;)

(d) 55
when x=50

TR = P*Q
TR = 80x-.5x^2
MR = 80-x

C = 100+30x
MC = 30

set MR = MC

80-x = 30
x = 50 (quanitity)

price = 80-25x = 55
Click to expand...


Bingo, we have a winner! When marginal revenue equals marginal cost, profit is at a maximum.......

 
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