Find the Maximum Profit: C(q) = (1/6)q^3 - 30q^2 + 2,500q, q>=0

[Happy101]

I'm trying to figure out the maximum profit and plot the points on the x. Here's the information I have:
--------------------------
Product production cost (C)
Production quantity (q)
Profit (P)
Product price is 1500.

[math]C(q) = (1/6)q^3 - 30q^2 + 2,500q[/math]q ≥ 0
---------------------------
Here's what what I'm doing but I'm doing something wrong... what is it?

Profit (P) = Sales revenue (R) - Production cost (C) = (Price × Production quantity) - (C(q)) = [math](1,500 × q) - ((1/6)q^3 - 30q^2 + 2,500q)[/math]
[math]-(1/6)q^3 + 30q^2 - 1,000q[/math][math]-1/2q^2 + 60q - 1,000 = 0[/math]
x=20 and x=100
Maximum = 60

The above doesn't seem to be correct. What am I doing wrong?

 

[stapel]

I'm trying to figure out the maximum profit and plot the points on the x. Here's the information I have:
--------------------------
Product production cost (C)
Production quantity (q)
Profit (P)
Product price is 1500.

[imath]C(q) = \left(\frac{1}{6}\right)\,q^3 - 30q^2 + 2,500q[/imath]

q ≥ 0
---------------------------
Here's what what I'm doing but I'm doing something wrong... what is it?

Profit (P) = Sales revenue (R) - Production cost (C)

= (Price × Production quantity) - (C(q))

[imath]= (1,500 × q) - ((1/6)q^3 - 30q^2 + 2,500q)[/imath]

[imath]= -(1/6)q^3 + 30q^2 - 1,000q[/imath]

I agree, up to this point.

[imath]-1/2q^2 + 60q - 1,000 = 0[/imath]

What you have posted means the following:

[imath]\qquad \frac{-1}{2q^2} + 60q - 1000 = 0[/imath]

I *think* you actually mean the following:

[imath]\qquad P'(q) = -\frac{1}{2}\, q^2 + 60q - 1000[/imath]

[imath]\qquad \textrm{Setting the derivative equal to zero, we get:}[/imath]

[imath]\qquad -\frac{1}{2}\, q^2 + 60q - 1000 = 0[/imath]

x=20 and x=100

I *think* this means that you have found zeroes of the derivative, [imath]P'(q)[/imath], at [imath]q = 20[/imath] and [imath]q = 100[/imath]. (There is no [imath]x[/imath] in the given information.) Since [imath]P(q)[/imath] is a negative cubic, it is obvious where the maximum must lie.

Maximum = 60

Are you saying that the maximum value is "equal to 60", as you've posted, or perhaps that the maximum value occurs "at [imath]q = 60[/imath]"?

The above doesn't seem to be correct. What am I doing wrong?

On what basis have you determined your work to be incorrect?

Please be complete. Thank you!

Eliz.

 

[Happy101]

I agree, up to this point.

What you have posted means the following:

[imath]\qquad \frac{-1}{2q^2} + 60q - 1000 = 0[/imath]

I *think* you actually mean the following:

[imath]\qquad P'(q) = -\frac{1}{2}\, q^2 + 60q - 1000[/imath]

[imath]\qquad \textrm{Setting the derivative equal to zero, we get:}[/imath]

[imath]\qquad -\frac{1}{2}\, q^2 + 60q - 1000 = 0[/imath]

I *think* this means that you have found zeroes of the derivative, [imath]P'(q)[/imath], at [imath]q = 20[/imath] and [imath]q = 100[/imath]. (There is no [imath]x[/imath] in the given information.) Since [imath]P(q)[/imath] is a negative cubic, it is obvious where the maximum must lie.

Are you saying that the maximum value is "equal to 60", as you've posted, or perhaps that the maximum value occurs "at [imath]q = 60[/imath]"?

On what basis have you determined your work to be incorrect?

To answer your question, when I entered the number into the cost function it didn't equal 0...

Note that I've figured this out now. Thank you.

 

[stapel]

To answer your question, when I entered the number into the cost function it didn't equal 0...

Which number? Why are you returning to the cost function instead of continuing with the profit function?

 

[blamocur]

Is not this the same problem as in your earlier thread ?