1. the cost of running an assembly line can be modelled by the function $c(x)=0.3x^{2}-0.9x + 1.675$, where…

1. the cost of running an assembly line can be modelled by the function $c(x)=0.3x^{2}-0.9x + 1.675$, where $c(x)$ is the cost per hour in thousands of dollars and $x$ is the number of items produced per hour in thousands. the most economical production level occurs when 1500 items are produced. verify this using the appropriate calculations for rate of change in cost.
Answer
Explanation:
Step1: Find the derivative of the cost function
The cost function is $C(x)=0.3x^{2}-0.9x + 1.675$. Using the power - rule for differentiation ($\frac{d}{dx}(ax^{n})=nax^{n - 1}$), the derivative $C'(x)$ is: $C'(x)=\frac{d}{dx}(0.3x^{2}-0.9x + 1.675)=0.3\times2x-0.9=0.6x - 0.9$
Step2: Set the derivative equal to zero to find the critical points
The most economical production level occurs when the rate of change of cost is zero. So we set $C'(x) = 0$. $0.6x-0.9 = 0$ Add $0.9$ to both sides: $0.6x=0.9$ Divide both sides by $0.6$: $x=\frac{0.9}{0.6}=1.5$ (in thousands of items) Since $x = 1.5$ thousands of items means $x = 1500$ items, the most economical production level is 1500 items.
Step3: Verify using the second - derivative test
Find the second - derivative $C''(x)$. Differentiating $C'(x)=0.6x - 0.9$ with respect to $x$, we get $C''(x)=\frac{d}{dx}(0.6x - 0.9)=0.6>0$. Since $C''(x)>0$ when $x = 1.5$, the cost function $C(x)$ has a minimum at $x = 1.5$ (or 1500 items).
Answer:
The most economical production level is 1500 items as verified by setting the first - derivative of the cost function equal to zero and using the second - derivative test to confirm it is a minimum.