Page:Calculus Made Easy.pdf/137

Now

${\frac {d^{2}C}{dP^{2}}}=+{\frac {b(2c+2P)}{(c+P)^{4}}}$,

which is positive for all the values of $P$; hence $P=+{\sqrt {\dfrac {b}{a}}}-c$ corresponds to a minimum.

(5) The total cost per hour $C$ of lighting a building with $N$ lamps of a certain kind is

$C=N\left({\frac {C_{l}}{t}}+{\frac {EPC_{e}}{1000}}\right)$,

where $E$ is the commercial efficiency (watts per candle),

$P$ is the candle power of each lamp,

$t$ is the average life of each lamp in hours,

$C_{l}=$ cost of renewal in pence per hour of use,

$C_{e}=$ cost of energy per $1000$ watts per hour.

Moreover, the relation connecting the average life of a lamp with the commercial efficiency at which it is run is approximately $t=mE^{n}$, where $m$ and $n$ are constants depending on the kind of lamp.

Find the commercial efficiency for which the total cost of lighting will be least.

${\begin{aligned}{\text{ We have}}\;C&=N\left({\frac {C_{l}}{m}}E^{-n}+{\frac {PC_{e}}{1000}}E\right),\\{\dfrac {dC}{dE}}&={\frac {PC_{e}}{1000}}-{\frac {nC_{l}}{m}}E^{-(n+1)}=0\end{aligned}}$

for maximum or minimum.

$E^{n+1}={\frac {1000\times nC_{l}}{mPC_{e}}}\quad$ and $\quad E={\sqrt[{n+1}]{\frac {1000\times nC_{l}}{mPC_{e}}}}$.