Answer
(a) $P^{'}(q)=72 e^{-0.2q} -14.4q e^{-0.2q} $dollars per unit
(b) $q=\frac{72}{14.4}=5$ units
(c) $P(5) \approx \$132.44$
Work Step by Step
$P(q)=72qe^{-0.2q}$dollars
(a)
$P^{'}(q)=\frac{d(72qe^{-0.2q} )}{dq}$
$P^{'}(q)=[72\frac{d(q)}{dq}] e^{-0.2q} +72q\frac{d( e^{-0.2q} )}{dq}$
$P^{'}(q)=72 e^{-0.2q} +72q e^{-0.2q}(-0.2) $
$P^{'}(q)=72 e^{-0.2q} -14.4q e^{-0.2q} $dollars per unit
(b)
$P^{'}(q)=72 e^{-0.2q} -14.4q e^{-0.2q} $dollars per unit
Putting $P^{'}(q)=0$
$0=72 e^{-0.2q} -14.4q e^{-0.2q} $
Or
$72 e^{-0.2q} -14.4q e^{-0.2q}=0 $
$e^{-0.2q} ( 72 -14.4q )=0 $
$\frac{1}{e^{0.2q}}( 72 -14.4q )=0 $
$ 72 -14.4q =0\times e^{0.2q} $
$ 72 -14.4q =0$
$ -14.4q =-72$
$ 14.4q =72$
$q=\frac{72}{14.4}=5$ units
(c)
Put $q=5$ in
$P(q)=72qe^{-0.2q}$dollars
$P(5)=72(5)e^{-0.2(5)}=360 e^{-1}= \frac{360}{e^1}\approx132.44$dollars
