In considering medical insurance for a certain operation, let $X$ equal the amount (in dollars) paid for the doctor and let $Y$ equal the amount paid to the hospital. In the past, the variances have been $Var(X)=8100,Var(Y)=10000,Var(X+Y)=20000$. Due to increased expenses, it was decided to increase the doctor's fee by \$500 and increase the hospital charge $Y$ by 8%. Calculate the variance of $X+500+(1.08)Y$, the new total claim.
Assuming independence of $X$ and $Y$ we know that in the past: $$Var(X+Y)=a^2(8100)+b^2(10000)=20000$$ Without more information I am unsure how to solve for $a$ and $b$ and proceed with the next calculation.
If $X$ and $Y$ were independent, then $$\operatorname{Var}[X+Y] = \operatorname{Var}[X] + \operatorname{Var}[Y],$$ but this is not possible because that would imply $20000 = 10000 + 8100$. So right away we know that $X$ and $Y$ are dependent in some fashion: their covariance is nonzero and the formula is $$\operatorname{Var}[aX+bY] = a^2 \operatorname{Var}[X] + b^2 \operatorname{Var}[Y] + 2ab \operatorname{Cov}[X,Y].$$ In the case that $a = b = 1$, this becomes $$ \operatorname{Var}[X+Y] = \operatorname{Var}[X] + \operatorname{Var}[Y] + 2 \operatorname{Cov}[X,Y].$$ Substituting the values for the variances we were given above allows us to obtain the covariance between $X$ and $Y$.
Now to compute the variance of $X + 500 + 1.08 Y$, we note that $$\operatorname{Var}[X+500+1.08Y] = \operatorname{Var}[X + 1.08Y]$$ since $500$ is constant. Now since the covariance is known, we may return to the first formula with the choice $a = 1$, $b = 1.08$, $\operatorname{Var}[X] = 8100$, $\operatorname{Var}[Y] = 10000$, and $\operatorname{Cov}[X,Y] = ???$ to obtain the desired variance.