a. The maximum profit is when : P` ( x ) = 0
P` ( x ) = ( - 0.15 x²+ 600 x - 140,500 ) `
P` ( x ) = - 0.3 x + 600
- 0.3 x + 600 = 0
0.3 x = 600
x = 600 : 0.3 = 2,000
Answer: 2000 units must be sold on a monthly basis to maximize the profit.
b. P max = - 0.15 · 2000² + 600 · 2000 - 140,500 =
= - 600,000 + 1,200,000 - 140,500 = $459,500
Answer: The maximum profit is $459,500.