the price that a company charged for a computer accessory is given by the equation 100 - 10x² where x is the…

Explanation:

Step1: Define the profit function

The revenue $R(x)$ is the price per - unit times the number of units. The price per unit is $p(x)=100 - 10x^{2}$, and the number of units is $x$ (in millions). So, $R(x)=x(100 - 10x^{2})=100x-10x^{3}$. The cost function $C(x)$ is $10x$ (since it costs $10$ dollars to make each of the $x$ million units). The profit function $P(x)$ is $P(x)=R(x)-C(x)=100x - 10x^{3}-10x=- 10x^{3}+90x$.

Step2: Set up the equation

We know that when $x = 2$, $P(2)=100$. We want to find $x$ such that $P(x)=100$, so we set up the equation $-10x^{3}+90x = 100$. Rearrange it to get $-10x^{3}+90x - 100 = 0$. Divide through by $- 10$ to obtain $x^{3}-9x + 10 = 0$.

Step3: Factor the cubic equation

We know that $x - 2$ is a factor of $x^{3}-9x + 10$ since $x = 2$ is a root. Using polynomial long - division or synthetic division: [ \begin{align*} x^{3}-9x + 10&=(x - 2)(x^{2}+2x - 5) \end{align*} ]

Step4: Solve the quadratic equation

Set $x^{2}+2x - 5=0$. Use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a = 1$, $b = 2$, and $c=-5$. [ \begin{align*} x&=\frac{-2\pm\sqrt{2^{2}-4\times1\times(-5)}}{2\times1}\ &=\frac{-2\pm\sqrt{4 + 20}}{2}\ &=\frac{-2\pm\sqrt{24}}{2}\ &=\frac{-2\pm2\sqrt{6}}{2}\ &=-1\pm\sqrt{6} \end{align*} ] We take the positive root $x=-1+\sqrt{6}\approx - 1+2.45 = 1.45$ (since $x$ represents the number of accessories produced in millions and cannot be negative).

Answer:

A. 1.45 million