Demand function

[johnjones]

I'm confused, should I be using the quotient rule or implicit differentiation.

Q: If demand (x units) for an item that sells at p dollars per unit is
x = 250/[1+2p], find dp/dx at the price of $2. Based on this, how much should the price be reduced in order to sell one more item?

I could let f(x) = 250 and g(x) = 1 + 2p, then go:

f'(x) = (1+2p)(250)' - (1+2p)'(250)
= (1+2p)(0) - (2)(250) [p is a constant, so derivative is 1]
= 0 - 500
= -500.

Is what I just did correct in terms of finding dp/dx? :?:

 

[Daniel_Feldman]

x = 250/[1+2p]

I think we need to do some implicit stuff...unless I'm not seeing something.

x(1+2p)=250

x+2px=250

1+2(p+x(dp/dx))=0

1+2p+2x(dp/dx)=0

dp/dx=(-2p-1)/-2x=(2p+1)/2x

 

[stapel]

I could let f(x) = 250 and g(x) = 1 + 2p....

Why not just deal with the information they gave you?

. . . . .x = 250/(1 + 2p)

They ask for dp/dx, so solve the above for p in terms of x:

. . . . .x(1 + 2p) = 250

. . . . .x + 2px = 250

. . . . .2px = 250 - x

. . . . .p = (250 - x) / (2x) = 125/x - 1/2

Differentiating:

. . . . .dp/dx = -125/x²

Plug "2" in for "x" in "x = 250/(1 + 2p)" and solve for the required value of x. Then plug into the above.

Eliz.

 

[johnjones]

I could let f(x) = 250 and g(x) = 1 + 2p....

Why not just deal with the information they gave you?

. . . . .x = 250/(1 + 2p)

They ask for dp/dx, so solve the above for p in terms of x:

. . . . .x(1 + 2p) = 250

. . . . .x + 2px = 250

. . . . .2px = 250 - x

. . . . .p = (250 - x) / (2x) = 125/x - 1/2

Differentiating:

. . . . .dp/dx = -125/x²

Plug "2" in for "x" in "x = 250/(1 + 2p)" and solve for the required value of x. Then plug into the above.

Eliz.

Do you mean plug 2 into p or x? I'm given price = 2. :shock: thx

 

[stapel]

Do you mean plug 2 into p or x? I'm given price = 2.

Read the exercise. Which variable stands for price?

Eliz.

 

[johnjones]

I got 31.25 .

2 = 250/(1 + 2p).


so am I interchanging the variables x and p?

 

[stapel]

so am I interchanging the variables x and p?

What do you mean by "interchanging" the two different variables? They stand for different things, don't they?

I'm sorry, but I don't understand.

Eliz.

 

[johnjones]

I think I confused myself too much... I think after plugging "x" into the equation, I got 31.25 for p.. which seems a bit weird, but whatever. Thx for the help, I think I get it.

-125 (50)^-2 = -0.05

 

[Gene]

For what it is worth, You were close in your first post but you for got d(a/b) = (b*da-a*db)/b^2
so dx/dp = -500/(1+2p)^2
When p=2 dx/dp=-20
dx=1 so
dp=1/-20 = -.05