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?
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?
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.
Interesting -- I call this "vertex form."
$\displaystyle \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.