Question

A concession stand at a sports stadium sells 600 bags of peanuts (each game) if the price of each bag is $4.00. For each $0.25 increase in price, 30 fewer bags are sold. What should the price of each bag of peanuts be so that the stand maximizes its revenue? What is the maximum revenue?

Answer 1

Answer:

Price = $4.50

Maximum revenue = $2,430

Step-by-step explanation:

The concession stand sells 600 bags of peanut at $4.00 (600, $4.00).

If there is a $1 increase in price, the number of bags old will decrease by 120, which mens that 480 bags of peanuts will be sold at $5.00 (480, $5.00).

Tracing a linear relationship between price and quantity sold with the two given points:

$m = \frac{5-4}{480-600}\\ m=-\frac{1}{120}$

$P - P_0=m*(Q-Q_0)\\P-4.00=-\frac{1}{120} (Q-600)\\P= -\frac{1}{120}Q +9$

The revenue function is given by the price multiplied by the quantity sold:

$R=Q*P= (-\frac{1}{120}Q +9)*Q\\R=-\frac{1}{120}Q^2 +9Q$

The value of 'Q' for which the derivate of the revenue function is zero, is the output level for which revenue is maximum:

$R'=0=-\frac{2}{120}Q+9\\Q_{max} = 540\\P_{max} = -\frac{540}{120} +9\\P_{max} =\ 4.50$

The total revenue of 540 units at $4.50 per unit is:

$R = \ 4.50*540\\R=\ 2,430$