A computer dealer has found that his monthly revenue for selling
personal computers can be modeled by the function R(x) = −120x^2+
6000x + 24000.
His costs can be modeled by C(x) = 3600x + 14400
How many computers should he sell monthly to achieve the maximum
profit?

Question

A computer dealer has found that his monthly revenue for selling
personal computers can be modeled by the function R(x) = −120x^2+
6000x + 24000.
 His costs can be modeled by C(x) = 3600x + 14400
How many computers should he sell monthly to achieve the maximum
profit?

anonymous · Answer

profit = revenue - costs
 =  -120x^2 + 6000x + 24000 -  3600x + 14400
 P =  -120x^2 + 2400x + 9600

we need value of for maximum profit

differentiate dP/dx and equate to zero and solve

anonymous · Answer

dP/dx =-240x + 2400 = 0
   x = 10 
10 computers

anonymous · Answer

have you done calculus  jfry?

anonymous · Answer

there is another way to find x other than differentiating (calculus}
shall i show you how?