Quadratics--maximize profit

jones19

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Oct 29, 2007
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Hi I have a question which i am stuck on... ill first write the question than say where I am and someone could correct me that way. Thanks in advance

Question:

Jim sells pies as a fundraiser. Last year he sold the pies for 15 dollars each and sold 133. Jim and his friends used their math skills and found that for every two dollar decrese below 15 dollars, they would sell ten more pies. Their total cost to run this funraiser is C(x) = 19x = 1545 where 'x' represents the number of 2 dollar decreases.

State the corresponding revenue fuction:
I got [ -20x(squared) - 116x + 1995 ]

*PLZ correct me if I am wrong*

Determine the corresponding profit function
I got [ -20x(squared) - 135x + 450 ]

*PLZ CORRECT*

How many pies must they sell to maximize profit?

= -20x(squared) - 135x + 450
= -20(x^2 + 6.75x) + 450
= -20(x^2 + 6.75x +11.39 - 11.39) + 450
= -20(x^2 + 6.75x +11.39) -227.8 + 450
= -20(x^2 + 6.75x + 11.39) + 222.2
= -20(x + 3.375)^2 + 222.2

**Thats what I got but to me it duznt make sense.. someone plz correct me or If i am right which i doubt tell me. Thanks and the last question is:

Determine the price they should charge in order to break even. How many pies will they sell at the break even point(s). In what price range will they make a profit.

I havent attempted this yet because first i wanna make sure of the awnser above. could someone plz help me out with that it would really be appreciaed thanks...


And another question I have besides the one above is

The difference between two numbers is 7. Find the numbers if the result of adding their sum and their product is a minimum.

*I do not understand it at all*


Thanks in advance
 
What grade are you in, Sir Jones?

Can you solve this equation for x:
3x^2 - 20x - 7 = 0

Do you know what a derivative is?

Need to know where you're at...
 
jones19 said:
Hi I have a question which i am stuck on... ill first write the question than say where I am and someone could correct me that way. Thanks in advance

Question:

Jim sells pies as a fundraiser. Last year he sold the pies for 15 dollars each and sold 133. Jim and his friends used their math skills and found that for every two dollar decrese below 15 dollars, they would sell ten more pies. Their total cost to run this funraiser is

C(x) = 19x = 1545 <---fix this equation before we can answer
where 'x' represents the number of 2 dollar decreases.
 
Denis said:
What grade are you in, Sir Jones?

Can you solve this equation for x:
3x^2 - 20x - 7 = 0

Do you know what a derivative is?

Need to know where you're at...

yes i can solve that equation .. no I do not kno wat a derivative is
 
jones19 said:
Hi I have a question which i am stuck on... ill first write the question than say where I am and someone could correct me that way. Thanks in advance

Question:

Jim sells pies as a fundraiser. Last year he sold the pies for 15 dollars each and sold 133. Jim and his friends used their math skills and found that for every two dollar decrese below 15 dollars, they would sell ten more pies. Their total cost to run this funraiser is C(x) = 19x + 1545 where 'x' represents the number of 2 dollar decreases. <-- Are you sure? It has nothing to do with # of pies?
 
yes I am hundred percent sure..... thats why i think my profit function is not accurate and thats wats making my maximixing the profit equation provide me with a weird number as well...
 
Okay - so x = # of $2 decreases

suppose

n = # of pies sold

then

(n-133)/10 = x

p(n) = 15 - (n-133)/5 .....price function

p(x)= 15 - 2x -----each

revenue = n * p(x)=(10x+133) * (15 - 2x)

profit function = R - C ....and then continue...
 
x = number of $2 decreases

revenue function is (price)(number sold) ...

R(x) = (15 - 2x)(133 + 10x)

cost function is given ...

C(x) = 19x + 1545

Profit = Revenue - Cost

P(x) = (15 - 2x)(133 + 10x) - (19x + 1545)

P(x) = -20x<sup>2</sup> - 135x + 450

P(x) will be a max when x = -b/(2a) = 135/-40 = -3.375

(-3.375)($ -2) = $6.75 increase in price to maximize profit.
 
jones19 said:
Determine the corresponding profit function
I got [ -20x(squared) - 135x + 450 ]
Had you not typoed, we would have known that was correct!
So you have:
-20x^2 - 135x + 450; divide by -5:
4x^2 + 27x - 90
Set to 0, then take 1st derivative:
4x^2 + 27x - 90 = 0
8x + 27 = 0
x = -27/8 = -3 3/8 : as per Sir Skeeter

Don't be "affraid" of the term 1st derivative: google it; quite simple.
 
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