An electronics store sells 70 digital cameras per month at a price of $320 each. For each $20 decrease in price, about 5 more cameras per month are sold. Use the verbal model and quadratic function to determine how much the store should charge per camera to maximize monthly revenue.

the price to maximize revenue monthly = $300

Step-by-step explanation:

From the model revenue equation, we have;

Revenue = Price x Sales

Let R (x) represent revenue

Thus;

R (x) = (320 - 20x) * (70 + 5x)

R (x) = - 100x² + 200x + 22400

Since the coefficient of x² is negative, it will be a downward opening parabola and thus the vertex will be at a maximum. Hence, we can use axis of symmetry formula to find x. Which is;

x = - b/2a

x = - 200 / (-2 x 100)

x = - 200/-200

x = 1

Thus the price to maximize revenue monthly will be gotten by plugging in 1 for x into (320 - 20x)

So, Price = (320 - 20 (1))

Price = $300