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.

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)  =

In order to maximize the revenue, set

 .............(i)

 .............(ii)

Solving (i) and (ii),

4 x (i)    ⇒       272x - 32y = 400

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

                        264x        = 450

∴  

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

the solution is h=10 because you add 9 to each side of the equation.

t = 3

step-by-step explanation:

put t = 3 in this expression :

(3²-1)(3²-2)(3² - 3)(3² - 4) = 8×7×6×5 = 56×30 = 1680