Answer
The number of cherry trees which should be planted per acre to maximize the annual yield for the acre is $40$ and the maximum number of pounds of cherries per acre is $1600$.
Work Step by Step
The number of pounds of cherries $N$ depends on the number of cherry trees, x. In particular the number of pounds of cherries is the original number $50$, minus the number lost to overcrowding.
The number of pounds of cherries per tree is the original number of pounds of cherries, $50$, minus the decrease due to overcrowding.
$N\left( x \right)=50-x$
The maximum number of cherry trees is $30+x$.
The number of pounds of cherries $R$ for the agency is the number of pounds of cherries per tree times the number of cherry trees $30+x$.
$\begin{align}
& R\left( x \right)=\left( 50-x \right)\left( 30+x \right) \\
& =1500+20x-{{x}^{2}}
\end{align}$
Thus, the number of pounds of cherry is $R\left( x \right)=-{{x}^{2}}+20x+1500$.
Compare it with the standard equation of quadratic function $f\left( x \right)=a{{x}^{2}}+bx+c$
The value of a is $-1$ , b is $20$ and c is $1500$
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, $a=-1$ , $a<0$ , $R$ has a maximum.
Recall that the maximum value is at $x=\frac{-b}{2a}$.
Substitute $-1$ for a and $20$ for b
$\begin{align}
& x=\frac{-b}{2a} \\
& =\frac{-20}{-2} \\
& =10
\end{align}$
The maximum number of cherry trees is $30+x$.
Substitute $10$ for x.
Thus, the maximum number of cherry trees is $30+x=40$
And the maximum value is $f\left( \frac{-b}{2a} \right)$.
Substitute $10$ for x in $R\left( x \right)=-{{x}^{2}}+20x+1500$
$\begin{align}
& R\left( x \right)=-{{x}^{2}}+20x+1500 \\
& =-\left( 100 \right)+20\left( 10 \right)+1500 \\
& =1600
\end{align}$
Hence, the maximum number of cherry trees is $40$ and maximum number of pounds of cherry is $1600$
