Question

for the given cost function ( c(x) = 72900 + 300x + x^2 ) find:
a) the cost at the production level 1050
b) the derivative of the cost function (( c(x) )) at the production level 1050
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Explanation:

Part (a)

Step1: Identify the cost function

The cost function is given by \( C(x) = 72900 + 300x + x^2 \). We need to find the cost when \( x = 1050 \).

Step2: Substitute \( x = 1050 \) into the cost function

Substitute \( x = 1050 \) into \( C(x) \):
\[

$$\begin{align*} C(1050)&= 72900 + 300(1050) + (1050)^2\\ &= 72900 + 315000 + 1102500\\ &= 72900 + 315000 = 387900\\ 387900 + 1102500&= 1490400 \end{align*}$$

\]

Step1: Find the derivative of the cost function

The cost function is \( C(x) = 72900 + 300x + x^2 \). The derivative of a constant is 0, the derivative of \( ax \) is \( a \), and the derivative of \( x^n \) is \( nx^{n - 1} \). So,
\[
C'(x)=\frac{d}{dx}(72900)+\frac{d}{dx}(300x)+\frac{d}{dx}(x^2)=0 + 300+2x
\]

Step2: Substitute \( x = 1050 \) into the derivative

Substitute \( x = 1050 \) into \( C'(x) \):
\[
C'(1050)=300 + 2(1050)=300+2100 = 2400
\]

Answer:

The cost at the production level 1050 is \( \boldsymbol{1490400} \).

Part (b)