Answer
A) Revenue in $2010$: $153.9$ billion
Cost in $2010$: $97.1$ billion
B) $P(t) = 0.6t^2-4.8t+44.8$
C) The profit in $2010$: $56.8$ billion
Work Step by Step
Given $$\begin{aligned}
R(t) &= 3.1t^2-33.7t+180.9\\
C(t)&= 2.5t^2-28.9t+136.1
\end{aligned}$$ Part A
The year $2010$ corresponds to $t= 10$. The revenue and cost for IBM in $2010$ is given by: $$\begin{aligned}
R(10) &= 3.1\cdot 10^2-33.7\cdot10+180.9 =\$153.9 \ billion\\
C(10)&= 2.5\cdot 10^2-28.9\cdot 10+136.1= \$97.1\ billion
\end{aligned}$$ Part B
The profit function is the difference between the revenue function and the cost function. It is given by: $$\begin{aligned}
P(t) &= R(t)-C(t) \\
&=3.1t^2-33.7t+180.9-(2.5t^2-28.9t+136.1)\\
&= (3.1-2.5)t^2+(28.9-33.7)t+(180.9-136.1)\\
P(t)&= 0.6t^2-4.8t+44.8
\end{aligned}$$ Part C
Set $t= 10$ to find the profit in $2010$.
$$\begin{aligned}
P(10)&= 0.6\cdot 10^2-4.8\cdot 10+44.8=\$56.8\ billion
\end{aligned}$$
