Question

hw 8 - derivative rules section 2.3: problem 10 (1 point)
a company decides to begin making and selling computers. the price function is given as follows:
p = - 35x + 3800,
where x is the number of computers that can be sold at a price of p dollars per unit. additionally, the financial department has determined that the weekly fixed cost of production will be 3000 dollars with an additional cost of 100 dollars per unit.
(a) find the revenue function in terms of x.
r(x)=
(b) use the financial departments estimates to determine the cost function in terms of x.
c(x)=
(c) find the profit function in terms of x.
p(x)=
(d) evaluate p(x) at x = 250.
p(250)=

Explanation:

Step1: Recall revenue formula

Revenue $R(x)$ is price per unit times number of units. Given $p = - 35x+3800$, then $R(x)=x\cdot p=x(-35x + 3800)=-35x^{2}+3800x$.

Step2: Recall cost - function formula

The fixed cost is 3000 and variable cost is 100 per unit. So $C(x)=100x + 3000$.

Step3: Recall profit - function formula

Profit $P(x)=R(x)-C(x)$. Substitute $R(x)=-35x^{2}+3800x$ and $C(x)=100x + 3000$ into the formula: $P(x)=-35x^{2}+3800x-(100x + 3000)=-35x^{2}+3700x - 3000$.

Step4: Differentiate the profit function

Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, if $P(x)=-35x^{2}+3700x - 3000$, then $P'(x)=-70x+3700$.

Step5: Evaluate the derivative at $x = 250$

Substitute $x = 250$ into $P'(x)$: $P'(250)=-70\times250+3700=-17500 + 3700=-13800$.

Answer:

(A) $R(x)=-35x^{2}+3800x$
(B) $C(x)=100x + 3000$
(C) $P(x)=-35x^{2}+3700x - 3000$
(D) $P'(250)=-13800$