A farmer owns 50 papaya trees. Each tree produces 600 papayas in a year. For each additional tree planted in the orchard, the output of each tree?

Question

A farmer owns 50 papaya trees. Each tree produces 600 papayas in a year. For each additional tree planted in the orchard, the output of each tree (including the pre-existing ones) drops by 5 papayas. How many trees should be added to the existing orchard in order to maximize the total production of papayas

Let $x$ be the number of additional trees planted, so resultant number of total trees = $50+x$ and the output per tree = $600-5x$ so the Output function $f(x)=(50+x)(600-5x) = 30,000 + 350x - 5x^2$

$f'(x)= 575 - 10x \Rightarrow x=35 $

Answer 1

If $f(x)=30,000 + 575x - 5x^2$ was your output function, then it is correctly maximized at $x=57.5$. Since a quadratic is symmetric about its maximum, $f(57)=f(58)$, and either $57$ or $58$ papaya trees will maximize production.

However, as Paul notes in the comments, your output function has an error. It ought to be

$$f(x)=(50+x)(600-5x) = 30,000 + \color{Red}{350}x - 5x^2$$

which is maximized at $x=35$, a unique solution.

Answer 2

Since $f$ increases up to $x=57.5$ and then decreases you should evaluate $f(57)$ and $f(58)$ separately and select $x= 57$ or $x=58$ depending on whichever is larger. If $f(57) = f(58)$ either choice of $x$ will maximize production.