Demand is q=-p^2+33p+9 copies of a book sold per week when price is p dollars. Can you help determine what price the company should charge to get the largest revenue?

Question

I solved this as a Max Revenue problem and got x=0 and x=22 so the books should be sold for $22 each. IS THIS RIGHT?

Also, cost is C=9q+100 dollars to sell q copies of a book in a week. What price should the company charge to get the largest weekly profit? What is the max possible profit weekly profit and how can you be certain that the profit is maximized?
I got the profit function to be P=p(-p^2+33p+9)-9(-p^2+33p+9)+100. I simplified this to P=(-p^2+33+9)(p-9+100). I then took the derivative using the product rule to get P'=-2p+33(p-9+100)+(1)(-p^2+33p+9). I know I then need to set this equal to zero if it is right, I then have to factor and say which value maximizes profit. IS THIS RIGHT? Please help me to solve this part, I do not know how to solve P' equal to zero and factor

Answer 1

largest revenue is q*p

revenue=qp
dR/dq=d/dq (-p^3 + 33p^2 + 9p)=0

solve that equation.

I get (22+-22.3)/2= 22

Now, cost.

Profit= revenue-cost
Profit= qp -9q+100
dP/dp= p(dq/dP) + q -9 dq/dp
where dq/dP=(-2p+33) and q=-p^2+33p+9

setting to zero

0=-2p^2+33p-p^2+33p+9-9(-2p+33)
0=-3p^2+48p-33*9

check that, then use the quadratic equation to solve it.

Answer 2

I see I made at least one error, in the constant. Check it line by line.

Answer 3

For the first part I took R' and set this equal to zero to get 22, is this right too.

Answer 4

I am confused on how you did the cost part. I think I got P' but I don't see this in your work. I know I have to set P' equal to zero but I don't know if I have to factor or something first

Answer 5

Yes, p=22 for max R.

On the second, profit, just start with revenue-cost, both as a function of p, simplify it as as abest you can, then dProfit/dp=0 and solve for p. The factoring will take care of itself...you need to end up with a quadratic in standard form.

Answer 6

I did simplify P' to be P'=-2p+33(p-9+100)+1(-p^2+33p+9) now I have to set the derivative,P', equal to zero and I do not know how to solve it.

Answer 7

We did what Matt said but we do not know what to do next, our teacher told us to set the derivative equal to zero and then factor to get more than one answer and say which one maximizes the profit but we cannot solve P'=0

Answer 8

P'=-2p+33(p-9+100)+1(-p^2+33p+9)=0
I didn't check that to make certain the derivative is correct. But if it is, gather terms, and

-p^2+ (-2+33+33)p + 3300 +9=0

check that.

Now use the quadratic equation to find the values of p that solve it.