the management of a local pest control business has determined that their net profit $p$, in thousands of…

the management of a local pest control business has determined that their net profit $p$, in thousands of dollars, is approximately given by $p(x)=-2x^{2}+100x - 1070$, where $x$ is the number of salespeople. using this equation, answer the following prompts: a) the minimum salespeople required for profitability is: b) the maximum salespeople possible for profitability is: c) the minimum salespeople required for a profit of 124 thousand dollars is: d) the maximum salespeople possible for a profit of 124 thousand dollars is: now find the maximum profit possible, the number of salespeople required, and rewrite the profit equation $p(x)$ in vertex form: maximum profit = thousand dollars required salespeople = $p(x)=$ question help: video check answer

the management of a local pest control business has determined that their net profit $p$, in thousands of dollars, is approximately given by $p(x)=-2x^{2}+100x - 1070$, where $x$ is the number of salespeople. using this equation, answer the following prompts: a) the minimum salespeople required for profitability is: b) the maximum salespeople possible for profitability is: c) the minimum salespeople required for a profit of 124 thousand dollars is: d) the maximum salespeople possible for a profit of 124 thousand dollars is: now find the maximum profit possible, the number of salespeople required, and rewrite the profit equation $p(x)$ in vertex form: maximum profit = thousand dollars required salespeople = $p(x)=$ question help: video check answer

Answer

Explanation:

Step1: Set profit equation for profitability

For profitability, $P(x)>0$. So, $- 2x^{2}+100x - 1070>0$. Divide by - 2 and reverse the inequality sign: $x^{2}-50x + 535<0$.

Step2: Use quadratic formula for roots

The quadratic formula for $ax^{2}+bx + c = 0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For $x^{2}-50x + 535 = 0$, $a = 1$, $b=-50$, $c = 535$. Then $x=\frac{50\pm\sqrt{(-50)^{2}-4\times1\times535}}{2\times1}=\frac{50\pm\sqrt{2500 - 2140}}{2}=\frac{50\pm\sqrt{360}}{2}=\frac{50\pm6\sqrt{10}}{2}=25\pm3\sqrt{10}\approx25\pm9.49$. The roots are $x_1=25 - 9.49 = 15.51$ and $x_2=25 + 9.49=34.49$. The minimum number of sales - people for profitability is 16 (since $x$ is a non - negative integer and $P(x)>0$ for $16\leq x\leq34$).

Step3: Set profit equation for $P(x)=124$

Set $-2x^{2}+100x - 1070 = 124$. Rearrange to get $-2x^{2}+100x-1194 = 0$. Divide by - 2: $x^{2}-50x + 597=0$.

Step4: Use quadratic formula again

For $x^{2}-50x + 597 = 0$, $a = 1$, $b=-50$, $c = 597$. Then $x=\frac{50\pm\sqrt{(-50)^{2}-4\times1\times597}}{2\times1}=\frac{50\pm\sqrt{2500 - 2388}}{2}=\frac{50\pm\sqrt{112}}{2}=\frac{50\pm4\sqrt{7}}{2}=25\pm2\sqrt{7}\approx25\pm5.29$. The roots are $x_1=25 - 5.29 = 19.71$ and $x_2=25 + 5.29 = 30.29$. The minimum number of sales - people for a profit of 124 thousand dollars is 20.

Step5: Rewrite in vertex form

The vertex form of a quadratic function $y = ax^{2}+bx + c$ is $y=a(x - h)^{2}+k$, where $h=-\frac{b}{2a}$ and $k = y(h)$. For $P(x)=-2x^{2}+100x - 1070$, $a=-2$, $b = 100$, $c=-1070$. $h=-\frac{100}{2\times(-2)} = 25$. $P(25)=-2\times(25)^{2}+100\times25-1070=-2\times625 + 2500-1070=-1250+2500 - 1070=170$. So, $P(x)=-2(x - 25)^{2}+170$.

Answer:

a) 16 b) 34 c) 20 d) 30 Maximum profit = 170 thousand dollars Required salespeople = 25 $P(x)=-2(x - 25)^{2}+170$