Suppose revenue is R(x)=1200x-100x cubed and cost is C(x)=600x+100 where x is number of widgets made.?

Suppose revenue is R(x)=1200x-100x cubed and cost is C(x)=600x+100 where x is number of widgets made. What is the value of x that maximizes. Assume x is greater than or equal to 0.


A) Revenue? B) Profit?

Suppose revenue is R(x)=1200x-100x cubed and cost is C(x)=600x+100 where x is number of widgets made.?
A) Revenue is maximum when R(x) is maximum. x=2


b) Profit is maximum when R(x)-C(x) is maximum. x=1
Reply:R(x) = -100x³ + 1200x


R'(x) = -300x² + 1200


R'= 0   ⇒   300x² = 1200


x² = 4


x=2 maximizes revenue





profit = revenue - cost


P(x) = R(x) - C(x)


= (1200x - 100x³) - (600x + 100)


= 1200x - 100x³ - 600x - 100


= -100x³ + 600x - 100





P'(x) = -300x² + 600


P' = 0   ⇒   300x² = 600


x² = 2


x = √2 ≅ 1.414, but you can't sell a partial widget so


x = 1 maximizes profit





∴ Do not quit your day job
Reply:A)





R(x) = 1200x - 100x³





Differentiate to get max value of x





dR/dx = 1200 -300x²


For max value dR/dx =0


0=1200 - 300x²


300x²=1200


x²=4


x=2








B) Profit = Revenue - Cost





Profit = 1200x - 100x³ - 600x - 100


Profit = 600x - 100x³ - 100





Differentiate





Dp/dx = 600 - 300x²


For max value dP/dx = 0


600-300x²=0


300x²=600


x²=2


x=√2





The number of item = 1