1. if the price of each ticket to a concert is $16, a total of 4000 tickets will be sold. for each $1…

Explanation:

Step1: Define revenue function

Let $x$ be the ticket - price. The number of tickets sold $n(x)=4000 - 100(x - 16)=5600-100x$. The revenue function $R(x)=x\cdot n(x)=x(5600 - 100x)=5600x-100x^{2}$.

Step2: Find the derivative

Differentiate $R(x)$ with respect to $x$. Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $R'(x)=\frac{d}{dx}(5600x-100x^{2})=5600 - 200x$.

Step3: Set the derivative equal to zero

To find the critical points, set $R'(x) = 0$. So, $5600-200x = 0$. Then, $200x=5600$, and $x = 28$.

Answer:

The ticket price should be set at $$28$ to maximize revenue.

Explanation for second part:

Step1: Find the derivative of the function

Given $f(x)=x^{2}-3x + 2$, using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we get $f'(x)=2x-3$.

Step2: Find the critical points

Set $f'(x)=0$. So, $2x - 3=0$, which gives $x=\frac{3}{2}$.

Step3: Evaluate the function at the critical point and the endpoint

Evaluate $f(x)$ at $x=\frac{3}{2}$: $f(\frac{3}{2})=(\frac{3}{2})^{2}-3\times\frac{3}{2}+2=\frac{9}{4}-\frac{9}{2}+2=\frac{9 - 18 + 8}{4}=-\frac{1}{4}$. Evaluate $f(x)$ at $x=-1$: $f(-1)=(-1)^{2}-3\times(-1)+2=1 + 3+2=6$.

Answer:

The absolute minimum value of $f(x)$ is $-\frac{1}{4}$ at $x = \frac{3}{2}$, and the absolute maximum value of $f(x)$ is $6$ at $x=-1$.