SOLUTION: The cost of producing x items (in hundreds) can be modeled by C (x)= 1/20x^2-4/5x+9 1/5 where C is the cost in thousands of dollars. how many items would you produce to minimize

Question 383066: The cost of producing x items (in hundreds) can be modeled by C (x)= 1/20x^2-4/5x+9 1/5
where C is the cost in thousands of dollars. how many items would you produce to minimize the cost and what would that cost be?
Answer by Jk22(389) (Show Source): You can put this solution on YOUR website!
The price is : C (x)= 1/20x^2-4/5x+9 1/5 thousand $ for x hundred items.

TO find the minimal price, need to find the cost where it's variation is zero, and for which any change would imply an increase in price, hence the variation is positive in price for any variation in item numbers :

Price variation (derivative) : C'(x) = 1/10x-4/5 = 0 => x = 8 hundred items

Variation of the latter : C''(x)=1/10 > 0 => hence any variation at x=8 imply an increase in the cost, this minimal cost being : C(8)=64/20-32/5+9+1/5=(16-32+45+1)/5=30/5=6 thousand $.

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