Question

A store has been selling 100 Blu-ray disc players a week at $600 each. A market survey indicates that for each $40 rebate offered to buyers, the number of units sold will increase by 80 a week. Find the demand function and the revenue function. How large a rebate should the store offer to maximize its revenue

Answer 1

Answer:

The rebate should be $220

Step-by-step explanation:

Demand Curve

It's the relationship between price (P) and quantity (Q) demanded a certain product or service.

(a) We need to find the function that relates both magnitudes assuming a linear equation. The equation of a line can be found with the point-point formula:

$\displaystyle Q-Q_1=\frac{Q_2-Q_1}{P_2-P_1}(P-P_1)$

Two sets of data are given: 100 Blu-ray disc players are sold a week at $600 each. The ordered pair for this condition is (P,Q)=(600,100).

The other point comes from the market survey: The number of units sold will increase by 80 (100+80=180) when the price goes down $40 (600-40=560). The new point is (P,Q)=(560,180)

We set up the equation of the demand

$\displaystyle Q-100=\frac{180-100}{560-600}(P-560)$

Rearranging

$-40Q+4000=80P-44800$

Or

$80P+40Q=48800$

Simplifying

$2P+Q=1120$

(b) The revenue function is Q times the price

$R=Q.P$

Solving the equation of the demand for P

$\displaystyle P=\frac{1120-Q}{2}$

Thus, the revenue is

$\displaystyle R=Q\cdot \frac{1120-Q}{2}$

$\displaystyle R=\frac{1120Q-Q^2}{2}$

(c) To find the optimum value of the revenue, we take the derivative of R and equate to 0

$\displaystyle R'=\frac{1120-2Q}{2}=560-Q=0$

Solving

Q=560 units a week

For which the revenue is

$\displaystyle R=\frac{1120(560)-560^2}{2}$

$R=\ 156,800$

And the price is

$\displaystyle P=\frac{1120-560}{2}=280$

$P=\ 280$

The rebate should be $600-$280=$220