# SOLUTION: I am not sure where to ask this. This is not a quadratic equation, but here is the problem. A company uses the function C(x)= 0.2X^2-3.4x+150 to model the unit cost in dollars

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 Question 16537: I am not sure where to ask this. This is not a quadratic equation, but here is the problem. A company uses the function C(x)= 0.2X^2-3.4x+150 to model the unit cost in dollars for producing x bars. For what number of bars is the unit cost at its minimum? What is the unit cost at that level of production? I guess I want to solve for C but I don't see two equations here. By using the brute force method, I get the lowest cost to be at 27 bars and a cost of 7.555555555 repeating.Answer by rapaljer(4667)   (Show Source): You can put this solution on YOUR website!If the cost function for producing x bars is , then to find the cost per bar, you must divide this by x. Cost per bar= or If you are in a calculus class (probably Calculus for Businees Majors or Concepts of Calculus!), then you take the derivative of this. Deriv = Set derivative = zero, and solve for x: Multiply both sides by the LCD Divide by .2 Square root both sides, where x>0: x= sqrt(750)= 27.386 Unit cost= Unit cost (when x= 27 bars) = 7.55555. . . R^2 at SCC