Answer
a) $2(x-6)(20-x)$
b) Price: $\$ 13$
Maximum weekly profit: $\$ 98$
Work Step by Step
a) The total revenue $c$ can be found as:
$c=(x-6)[20-2(x-10)]=2(x-6)(20-x)$
or, $c=-2x^2+52x-240$
b) From part (a), we have
$f(x)=-2x^2+52x-240$
The maximum value occurs at $x=-\dfrac{b}{2a}=-\dfrac{52}{2(-2)}=\$ 13$
Thus, the maximum value of the function $f(x)$ at $x=13$ is:
$c(13)=(13-6)(30-13)$
This gives:
$c(13)=\$ 98$
Hence, the feeder price $\$ 13$ leads to the maximum weekly profit of $\$ 98$.
