Line m passes through the points (3.7) and (6, 12) while line n passes
1 answer:
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Answer:
C(x) = $850 - $350x
Monthly Profit = -500x² + 1550x - 850
x = $1.55
the largest possible monthly profit = $351.25
Step-by-step explanation:
Given:
Demand equation:
q = -500x + 1200
where, q is the number of users who log on per month
x is the log-on fee you charge
Site maintenance fee = $10 per month
High-volume access fee = $0.70 per log-on
Now,
C(x) = $10 + ( $0.70 × q )
or
C(x) = $10 + ( $0.70 × (-500x + 1200) )
or
C(x) = $10 + ( - $350x + 840 )
or
C(x) = $850 - $350x
Total monthly income = qx
or
Total monthly income = ( -500x + 1200 ) × x
or
Total monthly income = -500x² + 1200x
now,
Profit = Income - cost
or
Monthly Profit = -500x² + 1200x - ( 850 - 350x )
or
Monthly Profit, P(x) = -500x² + 1550x - 850
For the largest possible monthly profit
= 0
or
=0
or
-2 × 500x + 1550 = 0
or
-1000x + 850 = 0
or
x = $1.55
Now,
the largest possible monthly profit will be at x = $1.55
substitute in the function of profit
P(1.55) = ( -500× 1.55² ) + ( 1550 × 1.55 ) - 850
or
the largest possible monthly profit = - 1201.25 + 2402.5 - 850
or
the largest possible monthly profit = $351.25