Assume that $C(x)$ is in dollars and $x$ is the number of units produced and sold. For the total-cost function $C(x)=0.01 x^{2}+0.6 x+40$, find $\Delta C$ and $C^{\prime}(x)$ when $x=90$ and $\Delta x=1$. \[ \Delta \mathrm{C}=\$ \] (Simplify your answer. Type an integer or decimal rounded to two decimal places as needed.)
Solution
Step 1 ：Given the total-cost function $C(x)=0.01 x^{2}+0.6 x+40$, we are asked to find the change in cost ($\Delta C$) and the derivative of the cost function ($C'(x)$) when $x=90$ and $\Delta x=1$.
Step 2 ：To find $\Delta C$, we evaluate the cost function at $x + \Delta x$ and subtract the cost function at $x$. So, $\Delta C = C(x + \Delta x) - C(x)$.
Step 3 ：Substituting $x=90$ and $\Delta x=1$ into the equation, we get $\Delta C = C(90 + 1) - C(90) = 177.41 - 175 = 2.41$.
Step 4 ：To find $C'(x)$, we take the derivative of the cost function. The derivative of $C(x)$ is $C'(x) = 0.02x + 0.6$.
Step 5 ：Substituting $x=90$ into the equation, we get $C'(90) = 0.02*90 + 0.6 = 2.4$.
Step 6 ：So, the change in cost $\Delta C$ when $x=90$ and $\Delta x=1$ is $\$2.41$ and the derivative of the cost function $C'(90)$ is $2.4$.
Step 7 ：Final Answer: \(\boxed{\Delta C = \$2.41}\) and \(\boxed{C'(90) = 2.4}\)
