The pressure in an ideal gas is given by $PV = RT$, where $T$ is the temperature of the gas, $V$ is the volume and $R$ is a constant.
Determine the maximum possible percentage error in $P$ when $T$ and $V$ are measured to within an accuracy of $6\%$ and $5\%$ respectively.
My attempt so far:
From the ideal gas law: $P = P(T,V) = \dfrac{RT}{V}$. The differential $\mathrm{d}P$ is given by $$ dP = \dfrac{\mathrm{d}P}{\mathrm{d}T}dT + \dfrac{\mathrm{d}P}{\mathrm{d}V}\mathrm{d}V = \dfrac{R}{V}\mathrm{d}T - \dfrac{RT}{V^2}\mathrm{d}V \;, $$ so $$ \dfrac{\mathrm{d}P}{P} = \dfrac{R}{VP}\mathrm{d}T - \dfrac{RT}{V^2P}\mathrm{d}V = \dfrac{R}{V(RT/V)}\mathrm{d}T - \dfrac{RT}{V^2(RT/V)}\mathrm{d}V = \dfrac{\mathrm{d}T}{T} - \dfrac{\mathrm{d}V}{V} $$ and $$ \left|\dfrac{\mathrm{d}P}{P}\right| = \left|\dfrac{\mathrm{d}T}{T} - \dfrac{\mathrm{d}V}{V}\right| \leq \left|\dfrac{\mathrm{d}T}{T}\right| + \left|\dfrac{\mathrm{d}V}{V}\right| = 6\% + 5\% = 11\% \;. $$ After this I draw a blank.
Could someone please fill in the blank and show me what I need to do next? Thanks.