SOLUTION: The marginal cost (C) of manufacturing x cell phones (in thousands) is given by: {{{C(x)= 5x^2-200x+4000}}} A. How many cell phones should be manufactured to minimize the marginal

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: The marginal cost (C) of manufacturing x cell phones (in thousands) is given by: {{{C(x)= 5x^2-200x+4000}}} A. How many cell phones should be manufactured to minimize the marginal      Log On


   

Question 1098121: The marginal cost (C) of manufacturing x cell phones (in thousands) is given by: C%28x%29=+5x%5E2-200x%2B4000
A. How many cell phones should be manufactured to minimize the marginal cost?
B. What is the minimum marginal cost?
My work was =
A. Using -b%2F2a = %28-200%29%2F%282%29%285%29= 20 or 20,000 phones to minimize the marginal cost.
B. I subbed in 20 as the x value in the given equation resulting in the minimum marginal cost being $2000.
I am uncertain if this is how to approach the question.

Found 3 solutions by josgarithmetic, Boreal, ikleyn:
Answer by josgarithmetic(39617) About Me  (Show Source):
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
Yes, your approach is fine.
-b/2a is the approach to get the vertex of the minimum, and it does equal 20, or 20,000 phones.
5(20^2)-200(20)+4000=2000-4000+4000=$2000
graph%28300%2C300%2C-10%2C30%2C-1000%2C10000%2C5x%5E2-200x%2B4000%29

Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.
The solution and the answer  "10"  given by @gosgarithmetic is   W R O N G.


For your safety,   S I M P L Y   I G N O R E   I T.