Riddle question max profit + units

[MrNeWBiE]

riddle question " max profit + units "

Homework Statement

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?

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 ?

 

[willem2]

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?

 

[MrNeWBiE]

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

or there is a better way ,,,

when 20$ ==> P = 20(x+0).(x)-(12x+200) " is it the right formula " ?

 

[MrNeWBiE]

well wait ,,,,,

this is the right formula ==> (20-x)(500x)-(12x+200)

please need your helps,,,,,

 

[MrNeWBiE]

well wait ,,,,,

this is the right formula ==> (20-x)(500x)-(12x+200)

please need your helps,,,,,

 

[Mark44]

Homework Statement

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?

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)

These formulas are applicable only if you are trying to find the maximum point on a parabola.

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

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

well wait ,,,,,

this is the right formula ==> (20-x)(500x)-(12x+200)

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.

 

[MrNeWBiE]

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)

 

[Mark44]

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).

 

[MrNeWBiE]

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 ?

 

[MrNeWBiE]

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

 

[Mark44]

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


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

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.

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


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

R = D * price

 

[MrNeWBiE]

so it's going to be

y=((-1/500)x+20).x-(12x+200)

then solve for it ? or something missing ?

 

[Mark44]

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).

 

[MrNeWBiE]

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 can't use " x " because "x" for the unit

so what i should put ,,, ?

 

[Mark44]

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?

 

[MrNeWBiE]

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 ..?

 

[Mark44]

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 ..?

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.

 

[MrNeWBiE]

d(f) = -500x+20

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

 

[MrNeWBiE]

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,,,,

 

[Mark44]

d(f) = -500x+20

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

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).

 

[Mark44]

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,,,,

It all depends on which variable is the independent variable and which is the dependent variable. If you want demand d as a function of price p, then the given points are (20, 0) and (19, 500).

 

[MrNeWBiE]

It all depends on which variable is the independent variable and which is the dependent variable. If you want demand d as a function of price p, then the given points are (20, 0) and (19, 500).

well i will use this because its give me the answers ,,, and today i should give the doctor the homework and yet 6 questions i didn't finish xD



thx mark ,,, i will try the other way later

 

[MrNeWBiE]

and btw the mark ,,,, since the question say " x " is the unit

then the first two points you gave me are right ,,,,

 

[Mark44]

If you write demand as a function of price, then each point on the graph will have coordinates (price, demand). Two points are (20, 0) and (19, 500).

If you want to look at price as a function of demand, then the order would be reversed, and the two functions would be inverses of each other.

 

[MrNeWBiE]

aha ,,,,

well i consider " demand as a function of price "

so i think it was right ,,,

thx a lot mark