6. (4 points) suppose that the manufacturer of a gas clothes dryer has found that when the unit price is $p$…
6. (4 points) 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 $r(p)=-4p^{2}+40p$. what unit price $p$ maximizes revenue? what is the maximum revenue?
Answer
Explanation:
Step1: Identify the function type
The revenue function $R(p)=-4p^{2}+40p$ is a quadratic function in the form $y = ax^{2}+bx + c$, where $a=-4$, $b = 40$, and $c = 0$.
Step2: Find the price that maximizes revenue
For a quadratic function $y=ax^{2}+bx + c$, the vertex of the parabola (which gives the maximum or minimum value depending on the sign of $a$) has its $x$-coordinate (in our case $p$-coordinate) at $p=-\frac{b}{2a}$. Substitute $a=-4$ and $b = 40$ into the formula: $p=-\frac{40}{2\times(-4)}=\frac{-40}{-8}=5$.
Step3: Find the maximum revenue
Substitute $p = 5$ into the revenue function $R(p)$: $R(5)=-4\times5^{2}+40\times5=-4\times25 + 200=-100 + 200=100$.
Answer:
The unit - price that maximizes revenue is $p = 5$ dollars, and the maximum revenue is $R = 100$ dollars.