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question 3
a company manufactures and sells baseball hats. theyve estimated the cost to manufacture h hats in a month, given by
c(h)=2.4h + 1960 dollars each month.
the demand for h hats at p dollars per hat is given by the demand equation
2h + 129p=6450.
at what price should the company sell these hats in order to maximize monthly profit? give your answer as a numerical value (no labels), rounded appropriately.

Explanation:

Step1: Express $H$ in terms of $p$ from the demand - equation

From $2H + 129p=6450$, we can solve for $H$:
$2H=6450 - 129p$, so $H = 3225-64.5p$.

Step2: Express the cost function in terms of $p$

Substitute $H = 3225-64.5p$ into the cost function $C(H)=2.4H + 1960$.
$C(p)=2.4(3225-64.5p)+1960=7740-154.8p + 1960=9700-154.8p$.

Step3: Express the revenue function in terms of $p$

The revenue function $R(p)=p\times H=p(3225 - 64.5p)=3225p-64.5p^{2}$.

Step4: Express the profit function in terms of $p$

The profit function $\pi(p)=R(p)-C(p)$.
$\pi(p)=(3225p-64.5p^{2})-(9700 - 154.8p)=-64.5p^{2}+3379.8p - 9700$.

Step5: Find the derivative of the profit function

The derivative of $\pi(p)$ with respect to $p$ is $\pi^\prime(p)=-129p + 3379.8$.

Step6: Set the derivative equal to zero to find the critical - points

Set $\pi^\prime(p)=0$, so $-129p + 3379.8 = 0$.
$129p=3379.8$, then $p=\frac{3379.8}{129}=26.2$.

Step7: Check the second - derivative to confirm it's a maximum

The second - derivative $\pi^{\prime\prime}(p)=-129<0$, so $p = 26.2$ is a point of maximum.

Answer:

$26.2$