the marginal cost ( c(x) ) (in dollars) of manufacturing ( x ) cell phones (in thousands) is given by:\n\n(…

Explanation:

Step1: Find the derivative of the cost function

The given cost function is (C(x) = 6x^{2}-230x + 8700). The derivative (C^{\prime}(x)) (using the power rule ((x^{n})^\prime=nx^{n - 1})) is (C^{\prime}(x)=12x-230).

Step2: Set the derivative equal to zero to find critical points

Set (C^{\prime}(x)=0), so (12x-230 = 0). Solving for (x): [ \begin{align*} 12x&=230\ x&=\frac{230}{12}=\frac{115}{6}\approx19.17 \end{align*} ] The second - derivative (C^{\prime\prime}(x) = 12>0), which means the function has a minimum at (x=\frac{115}{6}).

Step3: Calculate the number of cell - phones (for part (a))

Since (x) is in thousands, the number of cell - phones (N = 1000x). Substituting (x=\frac{115}{6}), we get (N=\frac{115\times1000}{6}=\frac{115000}{6}\approx19167)

Step4: Calculate the minimum marginal cost (for part (b))

Substitute (x = \frac{115}{6}) into the cost function (C(x)): [ \begin{align*} C\left(\frac{115}{6}\right)&=6\times\left(\frac{115}{6}\right)^{2}-230\times\frac{115}{6}+8700\ &=6\times\frac{13225}{36}-\frac{26450}{6}+8700\ &=\frac{13225}{6}-\frac{26450}{6}+8700\ &=\frac{13225 - 26450}{6}+8700\ &=\frac{- 13225}{6}+8700\ &=-\frac{13225}{6}+\frac{52200}{6}\ &=\frac{-13225 + 52200}{6}\ &=\frac{38975}{6}\approx6495.83 \end{align*} ]

Answer:

(a) (19167) cell - phones (b) ($6495.83)