Answer
A) Revenue in $2011$: $\$ 11,343.2$ million, profit in $2011$: $\$ 6160.8$ million
B) $C(t)= 76.2t^2-645t+3057.2$
C) Total cost: $\$5182.4$ million
Work Step by Step
Part A
Given $$\begin{aligned}
R(t) &= 164.1t^2-1353.9t+6380.0\\
P(t)&= 87.9t^2-708.9t+3322.8
\end{aligned}$$ The year $2011$ corresponds to $t= 11$. The revenue and profit for Pearson in $2011$ is given by: $$\begin{aligned}
R(11) &= 164.1\cdot 11^2-1353.9\cdot 11+6380.0\\
&=\$11,343.2 \ \text{million}\\
P(11)&= 87.9\cdot 11^2-708.9\cdot 11+3322.8\\
&= \$6160.8\ \text{million}
\end{aligned}$$ Part B
The profit function is the difference between the revenue function and the cost function. The cost function can be found from this relationship:
$$\begin{aligned}
P(t) &= R(t)-C(t) \\
C(t)&= R(t)-P(t)\\
&=164.1t^2-1353.9t+6380.0-(87.9t^2-708.9t+3322.8)\\
&= (164.1-87.9)t^2+(708.9-1353.9)t+(6380.0-3322.8)\\
C(t)&= 76.2t^2-645t+3057.2
\end{aligned}$$ Part C
Set $t= 11$ to find the cost in $2011$.
$$\begin{aligned}
P(10)&= 76.2\cdot 11^2-645\cdot 11+3057.2\\
&=\$5182.4\ \text{million}
\end{aligned}$$
