Functions. Using these formulas, find the optimal price

Question

A store can sell 225 DVDs per week at a price of $10 each. The manager estimates that for each $1 reduction in the price of the DVDs they will sell 30 more per week. The DVDs cost the store $5 each. If x stands for the number of $1 reductions in price, express the price P and the quantity sold Q in terms of x. Using these formulas, find the optimal price for the DVDs and the quantity that would be sold at that price.

Answer 1

x = number of $1 decreases

 

price = 10 - x

quantity = 225 + 30x

 

Revenue = (price)(quantity)

                 = (10 - x)(225 + 30x)

                 = -30x² + 75x + 2250

 

Maximum when x = -75/[2(-30)] = 75/60 = 1.25

 

Maximum revenue when the price = 10 - x = 10 - 1.25 = $8.75

 

Maximum revenue = -30(1.25)² + 75(1.25) + 2250 = $2296.88

 

quantity sold to maximize revenue = 225 + 30x = 225 + 30(1.25)

                                                                                    ≈263

Answer 2

The problem can be set up as follows

 

X >5 is the constraint

 

For every $1 drop in price, the number sold increases by 30.  Increase in sales is 30X

 

So the money realized is (225 + 30X)(10 - X) and X is chosen where this revenue is maximum

 

Here is the table that shows the plot of maximum at that point

 

0    2250
1    2295
2    2280
3    2205
4    2070
5    1875
6    1620

 

So the Revenue is maximum when X = 1.  However, the decrease is in terms of $1 per DVD and so the solution is X = 1

 

Cost of DVD is $9; Q sold is 225 + 30 = 255; Revenue is 2295