a small company shows the profits from their business with the function $p(x)= - 0.01x^{2}+60x - 500$, where…
a small company shows the profits from their business with the function $p(x)= - 0.01x^{2}+60x - 500$, where $x$ is the number of units they sell and $p$ is the profit in dollars.\na. how many units are sold by the company to earn the maximum profit?\nb. between which numbers of units sold does the company show a profit?\na. the company must sell $square$ units to earn the maximum profit.\n(round to the nearest whole number as needed.)
Answer
Explanation:
Step1: Identify the function type
The profit function $P(x)= - 0.01x^{2}+60x - 500$ is a quadratic function in the form $y = ax^{2}+bx + c$ where $a=-0.01$, $b = 60$ and $c=-500$.
Step2: Find the x - value of the vertex
For a quadratic function $y = ax^{2}+bx + c$, the x - value of the vertex (which gives the number of units for maximum profit) is $x=-\frac{b}{2a}$. Substitute $a=-0.01$ and $b = 60$ into the formula: $x=-\frac{60}{2\times(-0.01)}=\frac{60}{0.02}=3000$.
Step3: Find when the company shows a profit
The company shows a profit when $P(x)>0$, so we need to solve the inequality $-0.01x^{2}+60x - 500>0$. First, consider the corresponding quadratic equation $-0.01x^{2}+60x - 500 = 0$. Use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a=-0.01$, $b = 60$ and $c=-500$. $x=\frac{-60\pm\sqrt{60^{2}-4\times(-0.01)\times(-500)}}{2\times(-0.01)}=\frac{-60\pm\sqrt{3600 - 20}}{-0.02}=\frac{-60\pm\sqrt{3580}}{-0.02}$. $\sqrt{3580}\approx59.83$, so $x=\frac{-60\pm59.83}{-0.02}$. $x_1=\frac{-60 + 59.83}{-0.02}=\frac{-0.17}{-0.02}=8.5$ and $x_2=\frac{-60 - 59.83}{-0.02}=\frac{-119.83}{-0.02}=5991.5$. The company shows a profit when $9\leq x\leq5991$ (rounding to the nearest whole number).
Answer:
a. 3000 b. Between 9 and 5991 units.