a small tie shop finds that at a sales level of x ties per day its marginal profit is mp(x) dollars per tie, where mp(x)=1.25 + 0.02x - 0.0006x². also, the shop will lose $95 per day at a sales level of x = 0. find the profit from operating the shop at a sales level of x ties per day. p(x)=□

Question

Accepted Answer

# Explanation:
## Step1: Recall the relationship between profit and marginal profit
The marginal profit $MP(x)$ is the derivative of the profit function $P(x)$. So, $P(x)=\int MP(x)dx + C$.

## Step2: Integrate the marginal - profit function
Given $MP(x)=1.25 + 0.02x-0.0006x^{2}$.
$$
\begin{align*}
P(x)&=\int(1.25 + 0.02x-0.0006x^{2})dx+C\
&=\int1.25dx+\int0.02xdx-\int0.0006x^{2}dx + C\
&=1.25x+0.02\times\frac{x^{2}}{2}-0.0006\times\frac{x^{3}}{3}+C\
&=1.25x + 0.01x^{2}-0.0002x^{3}+C
\end{align*}
$$

## Step3: Determine the constant of integration $C$
We know that when $x = 0$, $P(0)=- 95$.
Substitute $x = 0$ and $P(0)=-95$ into $P(x)=1.25x + 0.01x^{2}-0.0002x^{3}+C$.
We get $P(0)=C=-95$

# Answer:
$P(x)=1.25x + 0.01x^{2}-0.0002x^{3}-95$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the relationship between profit and marginal profit

Step2: Integrate the marginal - profit function

Step3: Determine the constant of integration \(C\)

Answer: