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(4671) About Me  (Show Source):
You can put this solution on YOUR website!
If the cost function for producing x bars is
C%28x%29+=+0.2X%5E2-3.4x%2B150+, then to find the cost per bar, you must divide this by x.

Cost per bar= %280.2X%5E2-3.4x%2B150+%29%2Fx or 0.2x+-3.4+%2B150x%5E%28-1%29+
If you are in a calculus class (probably Calculus for Businees Majors or Concepts of Calculus!), then you take the derivative of this.
Deriv = .2+-150x%5E%28-2%29+

Set derivative = zero, and solve for x:
.2+-150x%5E%28-2%29+=+0
.2+-+150%2F%28x%5E2%29+=+0+
.2+=+150%2F%28x%5E2%29+

Multiply both sides by the LCD x%5E2
.2x%5E2+=+150+

Divide by .2
x%5E2+=+150%2F.2+=+750

Square root both sides, where x>0:
x= sqrt(750)= 27.386

Unit cost= %280.2X%5E2-3.4x%2B150+%29%2Fx
Unit cost (when x= 27 bars) = 7.55555. . .

R^2 at SCC