# Thread: minimizing marginal cost help

1. ## minimizing marginal cost help

the marginal cost c (in dollars) of manufacturing x cell phones in thousands is given by
C (x) = 5x^2-200x + 4000

A. how many cel phones should be manufactured to minimize the marginal cost ?

B. what is the minimum marginal cost?

2. Differentiate the function,

Set the derivative equal to 0 and solve for x.

Substitute this value of x into your function to get the minimum cost.

If you're not allowed to use a Calculus solution, then complete the square to get the function into turning point form. The turning point is the minimum value.

3. Originally Posted by Prove It
If you're not allowed to use a Calculus solution, then complete the square to get the function into turning point form. The turning point is the minimum value.
Interesting -- I call this "vertex form."

\begin{aligned}
C(x) &= 5x^2 - 200x + 4000 \\
&= 5(x^2 - 40x) + 4000 \\
&= 5(x^2 - 40x + 400 - 400) + 4000 \\
&= 5(x^2 - 40x + 400) - 2000 + 4000 \\
&= 5(x - 20)^2 + 2000
\end{aligned}

Since the x^2 term has a positive coefficient, the vertex, which is (20, 2000), is a minimum. Now you can answer the question.

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# minimizing marginal cost

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