Question

the revenue of a company can be modeled by $p = 0.0441t^2 - 0.7213t + 10.05$, where $t$ is the number of years since 1980 and $p$ is in millions of dollars. when will the revenue be 11 million? (note: make sure to round your answer down. if $t = 4.93$, it is still only 1984.) question help: ebook written example

Explanation:

Step1: Set up the equation

We know the revenue model is \( p = 0.0441t^{2}-0.7213t + 10.05 \), and we want to find \( t \) when \( p = 11 \). So we set up the equation:
\[ 0.0441t^{2}-0.7213t + 10.05=11 \]
Subtract 11 from both sides to get a quadratic equation in standard form:
\[ 0.0441t^{2}-0.7213t - 0.95 = 0 \]

Step2: Use the quadratic formula

For a quadratic equation \( ax^{2}+bx + c = 0 \), the quadratic formula is \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \). Here, \( a = 0.0441 \), \( b=- 0.7213 \), and \( c=-0.95 \).

First, calculate the discriminant \( D=b^{2}-4ac \):
\[

$$\begin{align*} D&=(-0.7213)^{2}-4\times0.0441\times(-0.95)\\ &= 0.52027369+4\times0.0441\times0.95\\ &=0.52027369 + 0.16758\\ &=0.68785369 \end{align*}$$

\]

Then, find the square root of the discriminant: \( \sqrt{D}=\sqrt{0.68785369}\approx0.8294 \)

Now, calculate the two solutions for \( t \):
\[

$$\begin{align*} t_1&=\frac{-(-0.7213)+\sqrt{D}}{2\times0.0441}\\ &=\frac{0.7213 + 0.8294}{0.0882}\\ &=\frac{1.5507}{0.0882}\\ &\approx17.58 \end{align*}$$

\]
\[

$$\begin{align*} t_2&=\frac{-(-0.7213)-\sqrt{D}}{2\times0.0441}\\ &=\frac{0.7213- 0.8294}{0.0882}\\ &=\frac{- 0.1081}{0.0882}\\ &\approx - 1.22 \end{align*}$$

\]

Since \( t \) represents the number of years since 1980, it cannot be negative. Also, we need to round down the positive solution. So we take \( t = 17 \) (because if we round down 17.58, we get 17).

Answer:

\( t = 17 \) (which means the year is \( 1980 + 17=1997 \))