Question

suppose that the manufacturer of a gas clothes dryer has found that, when the unit price is p dollars, the revenue r (in dollars) is r(p)= - 8p^2 + 24,000p. what unit price should be established for the dryer to maximize revenue? what is the maximum revenue? the unit price that should be established to maximize revenue is $ 1500. (simplify your answer.) the maximum revenue is $ (simplify your answer.)

Explanation:

Step1: Identify the quadratic - function form

The revenue function is $R(p)=-8p^{2}+24000p$, which is a quadratic function in the form $y = ax^{2}+bx + c$ where $a=-8$, $b = 24000$ and $c = 0$.

Step2: Find the price for maximum revenue

The vertex of a quadratic function $y=ax^{2}+bx + c$ has its $x$ - coordinate (in our case $p$ - coordinate) given by $p=-\frac{b}{2a}$. Substituting $a=-8$ and $b = 24000$ into the formula, we get $p=-\frac{24000}{2\times(-8)}=\frac{24000}{16}=1500$.

Step3: Calculate the maximum revenue

Substitute $p = 1500$ into the revenue function $R(p)=-8p^{2}+24000p$. So $R(1500)=-8\times(1500)^{2}+24000\times1500$. First, calculate $-8\times(1500)^{2}=-8\times2250000=-18000000$. Second, calculate $24000\times1500 = 36000000$. Then $R(1500)=-18000000 + 36000000=18000000$.

Answer:

$18000000$