A manufacturer of lighting fixtures has daily production costs of C = 800 − 30x + 0.25x^2, where C is the total cost (in dollars) and x is the number of units produced. How many fixtures should be produced each day to yield a minimum cost?
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C = 800 - 30x + 0.25x^2
Find the derivative:
C' = -30 + 0.5x
Set the derivative equal to 0 and solve for x:
-30 + 0.5x = 0
0.5x = 30
x = 60
Since the original equation is a quadratic with a positive coefficient on the x^2 term, and so the only critical point (x = 60) will be a minimum. Thus, there should be 60 fixtures produced each day to yield a minimum cost, which will be 800 - 30(60) + 0.25(60)^2 or -100 dollars, meaning they will profit 100 dollars a day by making 60 fixtures per day.
Find the derivative:
C' = -30 + 0.5x
Set the derivative equal to 0 and solve for x:
-30 + 0.5x = 0
0.5x = 30
x = 60
Since the original equation is a quadratic with a positive coefficient on the x^2 term, and so the only critical point (x = 60) will be a minimum. Thus, there should be 60 fixtures produced each day to yield a minimum cost, which will be 800 - 30(60) + 0.25(60)^2 or -100 dollars, meaning they will profit 100 dollars a day by making 60 fixtures per day.