Answer
The rental amount that will maximize the agency’s daily revenue is $\$35$ and the maximum daily revenue is $\$6125$.
Work Step by Step
The number of cars $N$ depends on the rental amount, x. In particular the number of cars is the original number $200$ , minus the number lost to rent increase.
The number of cars per day is the original number of cars, $200$, minus the decrease due to rent increase.
$N\left( x \right)=200-5x$.
The maximum rental amount is $30+x$.
The daily revenue $R$ for the agency is the number of cars times the rental amount $50+x$.
$\begin{align}
& R\left( x \right)=\left( 200-5x \right)\left( 30+x \right) \\
& =600+50x-5{{x}^{2}}
\end{align}$
Thus, the daily revenue is $R\left( x \right)=-5{{x}^{2}}+50x+600$.
Compare it with the standard equation of the quadratic function $f\left( x \right)=a{{x}^{2}}+bx+c$ .
The value of a is $-5$ , b is $50$ and c is $600$ .
Recall that if $a<0$ , then f has a maximum that occurs at $x=\frac{-b}{2a}$ and maximum value is $f\left( \frac{-b}{2a} \right)$
Since, $a=-5$ , $a<0$ , $R$ has a maximum.
Recall that the maximum value is at $x=\frac{-b}{2a}$
Substitute $-5$ for a and $50$ for b.
$\begin{align}
& x=\frac{-b}{2a} \\
& =\frac{-50}{-10} \\
& =5
\end{align}$
The maximum rental amount is $30+x$.
Substitute $5$ for x.
Thus, the maximum rental amount is $30+x=35$.
And the maximum value is $f\left( \frac{-b}{2a} \right)$.
Substitute $5$ for x in $R\left( x \right)=-5{{x}^{2}}+50x+600$.
$\begin{align}
& R\left( x \right)=-5{{x}^{2}}+50x+600 \\
& =-5\left( 25 \right)+50\left( 5 \right)+600 \\
& =6125
\end{align}$
Hence, the maximum rental amount is $\$35$ and maximum daily revenue is $\$6125$.
