## Precalculus (6th Edition) Blitzer

The number of walnut trees which should be planted per acre to maximize the annual yield for the acre is $25$ and the maximum number of pounds of walnuts per acre is $1250$.
The number of pounds of walnuts $N$ depends on the number of walnut trees, x. In particular the number of pounds of walnuts is the original number $60$, minus the number lost to overcrowding. The number of pounds of walnuts per tree is the original number of pounds of walnuts, $60$, minus the decrease due to overcrowding. $N\left( x \right)=60-2x$. The maximum number of walnut trees is $20+x$. The number of pounds of walnuts $R$ for the agency is the number of pounds of walnuts per tree times the number of walnut trees $20+x$. \begin{align} & R\left( x \right)=\left( 60-2x \right)\left( 20+x \right) \\ & =1200+20x-2{{x}^{2}} \end{align} Thus, the number of pounds of walnut is $R\left( x \right)=-2{{x}^{2}}+20x+1200$. Compare it with the standard equation of quadratic function $f\left( x \right)=a{{x}^{2}}+bx+c$ . The value of a is $-2$ , b is $20$ and c is $1200$ . 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, $a=-2$ , $a<0$ , $R$ has a maximum. Recall that the maximum value is at $x=\frac{-b}{2a}$. Substitute $-2$ for a and $20$ for b. \begin{align} & x=\frac{-b}{2a} \\ & =\frac{-20}{-4} \\ & =5 \end{align} The maximum number of walnut trees is $20+x$. Substitute $5$ for x. Thus, the maximum number of walnut trees is $20+x=25$. And the maximum value is $f\left( \frac{-b}{2a} \right)$ Substitute $5$ for x in $R\left( x \right)=-2{{x}^{2}}+20x+1200$. \begin{align} & R\left( x \right)=-2{{x}^{2}}+20x+1200 \\ & =-2\left( 25 \right)+20\left( 5 \right)+1200 \\ & =1250 \end{align} Hence, the maximum number of walnut trees is $25$ and maximum number of pounds of walnut is $1250$.