Question

Suppose that the price per unit in dollars of a cell phone production is modeled by p= $45 -0.0125x, where x is in thousands of phones produced, and the revenue represented by
thousands of dollars is R X .p. Find the production level that will maximize revenue.

Answer 1

Answer:

The production level that will maximize the revenue is 1800 (in thousands of phones produced), that is, production of 1,800,000 phones will maximize the revenue.

Step-by-step explanation:

The price per unit of phone is given as

p = 45 - 0.0125x

where x is in thousands of phones produced

Revenue = (price per unit) × (number of units)

Revenue = (45 - 0.0125x) × x

= (45x - 0.0125x²)

To find the maximum revenue, we need to obtain the maximum value of the revenue function.

R(x) = (45x - 0.0125x²)

At maximum point, (dR/dx) = 0 and (d²R/dx²) < 0

R(x) = (45x - 0.0125x²)

(dR/dx) = 45 - 0.025x

at maximum point, (dR/dx) = 0

(dR/dx) = 45 - 0.025x = 0

0.025x = 45

x = (45/0.025) = 1800

Hence, the production level that will maximize the revenue is 1800 (in thousands of phones produced)

That maximum revenue is thus

R(x) = (45x - 0.0125x²)

R(1800) = (45×1800) - (0.0125×1800²)

= 40,500

Hope this Helps!!!