# Math Help - i need to find largest revenue

1. ## i need to find largest revenue

the demand function given was q= -p(squared) + 33p +9 (18<- p <- 28) copies sold per week where price is p dollars. the question was determine what price the company should charge to obtain largest revenue. they also gave a cost function of c= 9p+100.. they wanted to know largest weekly profit also.

I tryed to do the rev pq and do derivatives but i kept coming up with squaring a negative..

Could someone help me out i'd greatly appreciate it!!

2. Originally Posted by maria
the demand function given was q= -p(squared) + 33p +9 (18<- p <- 28) copies sold per week where price is p dollars. the question was determine what price the company should charge to obtain largest revenue. they also gave a cost function of c= 9p+100.. they wanted to know largest weekly profit also.

I tryed to do the rev pq and do derivatives but i kept coming up with squaring a negative..

Could someone help me out i'd greatly appreciate it!!
So q is really the quantity sold per week. The product pq will give you the total revenue for price, p. To maximize the revenue, differentiate pq with respect to to q and set it equal to 0.

You should get in general $R'=P+Q\frac{dP}{dQ}$. Is this what you were doing?

Once you solve for that the way to find the weekly profit is pq-c.

3. ## more help!

I am pretty sure I understand, but i could be wrong

when i did the equation i did q = p(-q(squared) + 33.+9)

I multiplied p against the demand equation then took the derivative of that and set it equal to zero...

I ended up getting p(squared) = 33p + 9 but that doesnt seem right?

any suggestions?

4. Just in case a picture helps...

You may as well express R in terms of p then differentiate...

(I think Jameson meant differentiate with respect to p.)

Then solve of p when dR/dp = 0 (completing the square) ...

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5. Originally Posted by tom@ballooncalculus

(I think Jameson meant differentiate with respect to p.)
Yes you are correct. It seems to me usually the demand or quantity bought is a clear variable while the price is something a company can fully control, so it makes sense to use Q as the variable to differentiate with respect to. Since we were given a way to very nicely get Q for any P we choose, I see why P should be the variable in question.

6. I really appreciate the help

The only thing i guess im not getting is

Is the dr/dp

what i did was q = -3p(squared) + 66p+9 and set that equal to 0 to solve i think i am missing something in the equation

7. Originally Posted by maria
I really appreciate the help

The only thing i guess im not getting is

Is the dr/dp

what i did was q = -3p(squared) + 66p+9 and set that equal to 0 to solve i think i am missing something in the equation
No that looks correct. You have a quadratic equation in disguise. Think of "p" as "x" and notice you have an p^2 term, a p^1 term and a p^0 term. Use the quadratic formula to solve.

What you wrote is not Q, it is R', where R is the revenue function. In this problem everything is in terms of price, p, so we use that variable when differentiating. dR/dP is the derivative of the revenue equation with respect to p. You did that already.

8. THANK YOU!!!

I understood that completely i really appreciate the help! : )

9. question again...

i need some guidance to see if im doing this right

-p^3 + 33p + 9
First i did (pq) then i took the derivative to find revenue which i got
-3p^2 +33p + 9p = 0

well i keep getting to -p^2 = -11p -3

is this right? or where do i go from here?

Also i had to do a cost equation and had to find what -3p^2 + 18p -265 = 0

I go to p^2= -88.3 + 6p

Same question here, did i get the right numbers or of i did what do i do to solve?

thank you : )