## find expression for revenue, given demand function

Limits, differentiation, related rates, integration, trig integrals, etc.
stackey82
Posts: 4
Joined: Fri Mar 27, 2009 8:49 pm
Contact:

### find expression for revenue, given demand function

demand is given by q=(15000-1.5p)^(1/2) or the square root of (15000-1.5p)
In terms of the demand q, find an expression for the revenue

first I solved for p and got p= (15000-q^2)/1.5 then do I just multiply by the variable "q" and get (15000q-q^3)/1.5

I know that the answer is R(q)= (30000-2q^3)/3 so what am I doing wrong here?

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
demand is given by q=(15000-1.5p)^(1/2) or the square root of (15000-1.5p)
In terms of the demand q, find an expression for the revenue

first I solved for p and got p= (15000-q^2)/1.5 then do I just multiply by the variable "q" and get (15000q-q^3)/1.5
Since you need the revenue function R in terms of q, your first step is correct: solve the "demand" ("number of items sold") function for "p=". Since the revenue will be the product of the number of items sold and the per-item price, then R(q) = pq, where p is replaced with the expression you obtained.
I know that the answer is R(q)= (30000-2q^3)/3 so what am I doing wrong here?
I don't think you're doing anything wrong. I think there's a typo in the book, because I see no way of multiplying by q and not getting a "q" in each term of the function. The only other difference is that the solution writer "prettied up" the answer by getting rid of the decimal:

. . . . .$\left(\frac{15000q\, -\, q^3}{1.5}\right)\left(\frac{2}{2}\right)\, =\, \frac{30000q\, -\, 2q^3}{3}$

(Sorry for the tardy reply. I thought I'd answered this morning, but I guess a glitch "ate" my post.)

stackey82
Posts: 4
Joined: Fri Mar 27, 2009 8:49 pm
Contact:

### Re: find expression for revenue, given demand function

Thanks for your help. I thought I was going crazy and couldn't figure it out. Now I see what you mean "prettied it up." Once again thanks for your help its greatly appreciated.