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.

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!!!

answer: 1 + 1 = 11

step-by-step explanation: