the cost of manufacturing $x$ toasters in one day is given by $c(x)=-0.1x^{2}+20x - 310, 0 < x < 400$. (a)…
the cost of manufacturing $x$ toasters in one day is given by $c(x)=-0.1x^{2}+20x - 310, 0 < x < 400$. (a) list all the critical values of $c(x)$. if there are no critical values, enter none. $x =$ (b) find the interval(s) where $c(x)$ is increasing. (c) find the interval(s) where $c(x)$ is decreasing. (d) list the $x$-value(s) of all relative maxima of $c(x)$. if there are no relative maxima, enter none. $x =$ (e) list the $x$-value(s) of all relative minima of $c(x)$. if there are no relative minima, enter none. $x =$
Answer
Explanation:
Step1: Find the derivative
The derivative of $C(x)=- 0.1x^{2}+20x - 310$ using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$ is $C'(x)=-0.2x + 20$.
Step2: Find critical values
Set $C'(x)=0$. So, $-0.2x + 20 = 0$. Solving for $x$ gives $0.2x=20$, then $x = 100$. Since the domain is $0<x<400$, this critical value is valid.
Step3: Determine increasing and decreasing intervals
Choose test points in the intervals $(0,100)$ and $(100,400)$. For the interval $(0,100)$, let $x = 50$. Then $C'(50)=-0.2\times50 + 20=-10 + 20 = 10>0$, so $C(x)$ is increasing on $(0,100)$. For the interval $(100,400)$, let $x = 200$. Then $C'(200)=-0.2\times200+20=-40 + 20=-20<0$, so $C(x)$ is decreasing on $(100,400)$.
Step4: Find relative extrema
Since $C(x)$ changes from increasing to decreasing at $x = 100$, there is a relative maximum at $x = 100$ and no relative minimum.
Answer:
(a) $x = 100$ (b) $(0,100)$ (c) $(100,400)$ (d) $x = 100$ (e) NONE