How would you express revenue "R" as a quadratic function of price "p" when cost "C=90,000+30x"and x=9,000-30p

Suppose that in the market research example in this section the demand equation (1) is changed to x=9,000-30p and the cost equation (2) is changed to C=90,000+30x.





How would you express revenue R as a quadratic function of price p?





C as a linear function of price p is C=360,000-90p.

How would you express revenue "R" as a quadratic function of price "p" when cost "C=90,000+30x"and x=9,000-30p
Revenue = Price x Quantity: R = P * Q


Define C = Cost


Define Q = Quantity which you call X in your equations.





Cost is defined by the equations:


C = 360,000 - 90P


C = 90,000 + 30Q





put these equations together and





360,000 - 90P = 90,000 + 30Q





Solve for Q





270,000 - 90P = 30Q


9,000 - 3P = Q





Now, go back to the original Revenue Equation:


R = P * Q





R = P * (9,000 - 3P)


R = 9,000P - P^2 a quadratic function of price P





Note: if P = 0, R = 0. Then R increases until P = 4,500. This is the maximum R which you can find by differentiating R, setting = 0 and solving for P. After this point, if we keep increasing P, R coninues to fall back to 0 at P = 9,000.





Note: there is a little problem, here. Your original equations state:


C = 90,000 + 30X and X = 9,000 - 30P Substitute for X


C = 90,000 + 30 * (9,000 - 30P)


C = 90,000 + 270,000 - 900P


C = 360,000 - 900P





THIS IS DIFFERENT FROM YOUR EQUATION: C=360,000-90P


One equation says 900P, the other says 90P.





The math is the same. Depending on which is correct, your answer will change!


Good Luck!