Question

A concession stand at an athletic event is trying to determine how much to sell cola and iced tea for in order to maximize revenue. Let x be the price per cola and y the price per iced tea. Demand for cola is 100 – 34x + 5y colas per game and iced tea is 50 + 3x – 16y iced teas per game The concession stand should charge: dollars per cola, dollars per iced tea, in order to maximize revenue. The maximum revenue for one game is: dollars.

Answer 1

Solution :

Demand for cola : 100 – 34x + 5y

Demand for cola : 50 + 3x – 16y

Therefore, total revenue :

x(100 – 34x + 5y) + y(50 + 3x – 16y)

R(x,y)  = $100x-34x^2+5xy+50y+3xy-16y^2$

$R(x,y) = 100x-34x^2+8xy+50y-16y^2$

In order to maximize the revenue, set

$R_x=0, \ \ \ R_y=0$

$R_x=\frac{dR }{dx} = 100-68x+8y$

$R_x=0$

$68x-8y=100$  .............(i)

$R_y=\frac{dR }{dx} = 50-32x+8y$

$R_y=0$

$8x-32y=-50$  .............(ii)

Solving (i) and (ii),

4 x (i)    ⇒       272x - 32y = 400

     (ii)   ⇒   (-)     8x - 32y = -50  

                        264x        = 450

∴   $x=\frac{450}{264}=\frac{75}{44}$

     $y=\frac{175}{88}$

So, x ≈  $ 1.70      and    y = $ 1.99

    R(1.70, 1.99) = $ 134.94

Thus, 1.70 dollars per cola

          1.99 dollars per iced ted to maximize the revenue.

Maximum revenue = $ 134.94