Riddle question max profit + units

1. Apr 17, 2010

MrNeWBiE

riddle question " max profit + units "

1. The problem statement, all variables and given/known data

There is no demand for a new brand of mobile phones if the price is 20 $or more. For each drop of 1$ in the price, the demand increases by 500 units.The cost of producing x units is (12x + 2000)$. How many phones should be produced and sold to obtain a maximum profit? What is the price per unit charged to get maximum profit? 3. The attempt at a solution well to know the units i should use x=-b/2a and the max$ from y=c-(b^2/4a)

but can someone tell me how to make the equation for this question ,,,

P=R-C ,,,
C=(12x + 2000) ,,,
how to find R ?

2. Apr 17, 2010

willem2

Re: riddle question " max profit + units "

can you give a formula for what the net profit is, if the price is y. It only needs to be valid for (0<y<20)
can you find the maximum of that?

3. Apr 17, 2010

MrNeWBiE

Re: riddle question " max profit + units "

:uhh::uhh:

so i keep going down in the numbers till i find the max ,,,?

or there is a better way ,,, :uhh:

when 20$==> P = 20(x+0).(x)-(12x+200) " is it the right formula " ? 4. Apr 18, 2010 MrNeWBiE Re: riddle question " max profit + units " well wait ,,,,, this is the right formula ==> (20-x)(500x)-(12x+200) please need your helps,,,,, 5. Apr 18, 2010 MrNeWBiE Re: riddle question " max profit + units " well wait ,,,,, this is the right formula ==> (20-x)(500x)-(12x+200) please need your helps,,,,, 6. Apr 18, 2010 Mark44 Staff: Mentor Re: riddle question " max profit + units " These formulas are applicable only if you are trying to find the maximum point on a parabola. First, it's not a formula, because it is not an equation, and besides, you have an error in it. You need to approach this problem in a more systematic way, by finding the demand and revenue. You can assume that the number of phone units sold will be equal to the demand. What's the equation that represents the demand D as a function of the sale price? You can assume that the demand function is linear - IOW, its graph is a straight line. Two points on this line are (0, 20) and (500, 19). The revenue is going to be the number of units sold (the demand) times the sale price. 7. Apr 18, 2010 MrNeWBiE Re: riddle question " max profit + units " yet i didn't understand :zzz: ,,, how to find D and the price ,, ? because x is for the units not for$ ,,, so it's hard for me ,,,

the only think on my brain for it ,,,

(500x)(20-x)

8. Apr 18, 2010

Staff: Mentor

Re: riddle question " max profit + units "

Do you know how to find the equation of a straight line given two points on the line? The line goes through (0, 20) and (500, 19).

9. Apr 19, 2010

MrNeWBiE

Re: riddle question " max profit + units "

so i sould find the slope then solve it ,,, ?

y=(-1/500)x+20 ,,,, this is the demand

and and from where i find the price and put it ,,, ?

is it ,,,, R = D . price or R = D . units ?

Last edited: Apr 19, 2010
10. Apr 19, 2010

MrNeWBiE

Re: riddle question " max profit + units "

if it's R = D . units i think i solve it ,,,,

it's going to be 2000 units and 6000 $... or not all my work is wrong 11. Apr 19, 2010 Mark44 Staff: Mentor Re: riddle question " max profit + units " No. You have two points on the line, so find the slope. Then use either of the points and the slope in the point-slope form of the equation of a line. R = D * price 12. Apr 19, 2010 MrNeWBiE Re: riddle question " max profit + units " so it's going to be y=((-1/500)x+20).x-(12x+200) then solve for it ? or something missing ? 13. Apr 19, 2010 Mark44 Staff: Mentor Re: riddle question " max profit + units " No again. You keep jumping to incorrect conclusions without showing the work that you are doing. You have two points on the line, so find the slope. It is NOT -1/500! Show the work you do to find the slope of the line between (0, 20) and (500, 19). 14. Apr 19, 2010 MrNeWBiE Re: riddle question " max profit + units " the slope = 20-19/0-500 .... m = -1/500 ,,,, still the slope wrong ? y-y1=m(x-x1) ,,,,, y-20=-1/500(x-0) y=(-1/500)x+20 now plug it (-1/500)x+20 ( price ) - (12x+200) ,,,, what i put for the price ,,, i cant use " x " because "x" for the unit so what i should put ,,, ? 15. Apr 19, 2010 Mark44 Staff: Mentor Re: riddle question " max profit + units " Sorry, my mistake, and I apologize for confusing you. I wrote the points in the wrong order way back in post #8. The points on the demand function should be (20, 0) and (19, 500). The slope is (500 - 0)/(19 - 20) = -500. So now what do you get for the demand function? 16. Apr 19, 2010 MrNeWBiE Re: riddle question " max profit + units " i go back to my old book and i found demand function = price ..... where is R = price . unit .... so p = R - c ==> p = p.x-C ,,, right or i read something wrong ..? 17. Apr 19, 2010 Mark44 Staff: Mentor Re: riddle question " max profit + units " In this problem, demand is a specific function of price. The problem says that at$20, the demand is 0, and for each $1 reduction in price, the demand increases by 500 units. Per the given information, the demand function is linear. The slope of the line is -500 (units/$). What is the equation of the demand function?

Yes, revenue = price x number of units sold. In this problem the assumption is that the no. of units sold equals the demand.

This equation of yours -- p = R - c -- doesn't make any sense, if p represents price. The revenue is the amount brought in from the sale of some number of phones.
Profit = Revenue - Cost, but P (Profit) and p (unit price) are different.

18. Apr 19, 2010

MrNeWBiE

Re: riddle question " max profit + units "

d(f) = -500x+20

then i plug this with (cost) or i should add something before adding ?

19. Apr 19, 2010

MrNeWBiE

Re: riddle question " max profit + units "

yo mark ,,,, i think the first points you gave me was right ,,,
because i try with calculator every time 1\$ down ,,, add 500 units ,,,

and i get the same answer as the first two ponits,,,,

20. Apr 19, 2010

Staff: Mentor

Re: riddle question " max profit + units "

The slope is right, but the formula isn't. A better choice instead of x might be p, for price.

We know that d(20) = 0, but your formula gives d(20) = -500(20) + 20 = -9980. You can get the equation of the line by using one of the points you know, say (20, 0) and the slope, -500, the equation of the line is d - 0 = -500(p - 20).
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