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the cost of manufacturing $x$ toasters in one day is given by $c(x)=-0.…

Question

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. $x = 100$ (b) find the interval(s) where $c(x)$ is increasing. $(0,100)$ (c) find the interval(s) where $c(x)$ is decreasing. $(100,400)$ (d) list the $x$-value(s) of all relative maxima of $c(x)$. if there are no relative maxima, enter none. $x = 100$ (e) list the $x$-value(s) of all relative minima of $c(x)$. if there are no relative minima, enter none. $x = none$

Explanation:

Step1: Find the derivative

The derivative of $C(x)=- 0.1x^{2}+20x - 310$ is $C'(x)=-0.2x + 20$ using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$.

Step2: Find critical values

Set $C'(x)=0$. So, $-0.2x + 20=0$. Solving for $x$ gives $0.2x=20$, then $x = 100$.

Step3: Determine increasing interval

We test intervals using the first - derivative test. For the interval $(0,100)$, let's choose $x = 50$. Then $C'(50)=-0.2\times50 + 20=-10 + 20=10>0$. So $C(x)$ is increasing on $(0,100)$.

Step4: Determine decreasing interval

For the interval $(100,400)$, let's choose $x = 200$. Then $C'(200)=-0.2\times200+20=-40 + 20=-20<0$. So $C(x)$ is decreasing on $(100,400)$.

Step5: Find relative maxima and minima

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