Question

section 2.8: optimization (homework)
score: 20/120 answered: 2/12
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question 3
0/10 pts 4 99 details
the pear company sells pphones. the cost to manufacture x pphones is c(x)=-19x^{2}+47000x + 19227 dollars (this includes overhead costs and production costs for each pphone). if the company sells x pphones for the maximum price they can fetch, the revenue function will be r(x)=-26x^{2}+187000x dollars.
how many pphones should the pear company produce and sell to maximize profit? (remember that profit = revenue - cost.)
x=

Explanation:

Step1: Define the profit function

The profit function $P(x)$ is given by $P(x)=R(x)-C(x)$. Substitute $R(x)= - 26x^{2}+187000x$ and $C(x)=-19x^{2}+47000x + 19227$ into the profit - function formula:
$P(x)=(-26x^{2}+187000x)-(-19x^{2}+47000x + 19227)$
$P(x)=-26x^{2}+187000x + 19x^{2}-47000x-19227$
$P(x)=(-26 + 19)x^{2}+(187000 - 47000)x-19227$
$P(x)=-7x^{2}+140000x-19227$.

Step2: Find the derivative of the profit function

The derivative of $P(x)$ with respect to $x$ is $P^\prime(x)$. Using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have:
$P^\prime(x)=\frac{d}{dx}(-7x^{2}+140000x-19227)=-14x + 140000$.

Step3: Set the derivative equal to zero and solve for x

To find the critical points, set $P^\prime(x) = 0$.
$-14x+140000 = 0$.
Add $14x$ to both sides: $140000=14x$.
Divide both sides by 14: $x=\frac{140000}{14}=10000$.

Step4: Confirm it's a maximum

The second - derivative of $P(x)$ is $P^{\prime\prime}(x)=\frac{d}{dx}(-14x + 140000)=-14<0$. Since the second - derivative is negative, when $x = 10000$, the profit function $P(x)$ has a maximum.

Answer:

$10000$