College Algebra, Chapter 4, 4.1, Section 4.1, Problem 70

Suppose at a certain vineyard it is found that each grape vine produces about 10 pounds of grapes in a season when about 700 vines are planted per acre. For each additional vine that is planted, the production of the each vine decreases by about 1 percent. So the number of pounds of grapes produced per acre is modeled by

$A(n) = (700 + n)(10 - 0.01n)$

where $n$ is the number of additional vines planted. Find the number of vines that should be planted to maximize grape production.

We rewrite the function as


$\begin{equation} \begin{aligned} A(n) =& (700 + n) (10 - 0.01 n) \\ \\ A(n) =& 7000 - 7n+ 10n - 0.01n^2 \\ \\ A(n) =& 7000 + 3n - 0.01n^2 \end{aligned} \end{equation}$


The function $A$ is a quadratic function with $a = -0.01$ and $b = 3$. Thus, its maximum value occurs when

$\displaystyle n = - \frac{b}{2a} = - \frac{3}{2(-0.01)} = 150$ vines planted

The maximum production is $A(150) = 7000 + 3(150) - 0.01(150)^2 = 7225$.

It shows that $150$ vines should be planned in order to have a maximum production of $7225$ grapes.