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!!!
