Question 4.3.8: Petroleum Consumption The petroleum consumption C (in quads)...
Petroleum Consumption
The petroleum consumption C (in quads) in the United States from 1970–1995 can be modeled by the function
C(x) = 0.003x³ – 0.139x² + 1.621x + 28.995,
where x = 0 represents 1970, x = 1 represents 1971, and so on.
The model indicates that C(5) = 34 quads of petroleum were consumed in 1975. Find another year between 1975 and 1995 when the model indicates that 34 quads of petroleum were consumed. (The model was slightly modified from the data obtained from the Statistical Abstracts of the United States.)
We are given that
\begin{aligned}C(5) &=34 & & \\C(x) &=(x-5) Q(x)+34 & & \text { Division algorithm and Remainder Theorem } \\C(x)-34 &=(x-5) Q(x) & & \text { Subtract } 34 \text { from both sides. }\end{aligned}
Therefore, 5 is a zero of F(x)=C(x)-34 =0.003 x^{3}-0.139 x^{2}+1.621 x-5.005 .
Finding another year between 1975 and 1995 when the petroleum consumption was 34 quads requires us to find another zero of F(x)=C(x)-34 that is between 5 and 25. Because 5 is a zero of F(x), we use synthetic division to find the quotient Q(x).
\begin{matrix} \underline{5|} & \\ \\ \\ \\ \end{matrix} \begin{matrix} 0.003 & -0.139& 1.621 & -5.005 \\ & 0.015 & -0.62 & 5.005\\ \hline 0.003 &-0.124 & 1.001 & |0 \end{matrix}
We solve the depressed equation Q(x)=0.
\begin{array}{ll}0.003 x^{2}-0.124 x+1.001=0 & Q(x)=0.003 x^{2}-0.124 x+1.001 \\x=\frac{0.124 \pm \sqrt{(-0.124)^{2}-4(0.003)(1.001)}}{2(0.003)} & \text { Quadratic formula } \\x=11 \text { or } x=30 . \overline{3} & \text { Use a calculator. }\end{array}
Because x needs to be between 5 and 25, this model tells us that in 1981 (x=11), the petroleum consumption was also 34 quads.