Math Question Max-Min problem

Question

Determine how many apple trees should be planted per hectare to yield the greatest number of apples. If no more than 20 trees are planted per hectare, apple trees grown in Toronto produce 600 apples each year. The number of apples produced by a tree per hectare decreases by 15 for every additional tree.

Answer 1

Clearly, the optimal number of trees is at least 20 per hectare. If there were fewer than 20 trees, we could add more trees without decreasing the number of apples per tree.

So let's say the number of trees is t+20. Then the number of apples produced per tree is 600-15t; we start at 600 when t=0 (at 20 trees), and decrease by 15 per extra tree.

The total number of apples produced is (number of trees) times (number of apples per tree), which is now (t+20)(600-15t). We need to maximize this quantity.

Expand the product: 600t+12,000-15t²-300t=12,000+300t-15t². We can maximize this by completing the square.

First take out the -15: -15(t²-20t-800)

When a quadratic like is written as (x-a)², it expands as x²-2ax+a². In our case, -2a=-20, so a=10 and a²=100

Rewrite the expression as -15(t²-20t+100-100-800)=-15((t-10)²-900)=13,500-15(t-10)²

Remember, we're trying to pick t (20 less than the number of trees) to maximize this expression for the number of apples produced. Since (t-10)² is always positive, we've expressed the number of apples as 13,500 minus (a positive number); clearly, we want to subtract 0 from the amount of apples, so we set -15(t-10)²=0 and solve.

(t-10)²=0

t-10=0

t=10, so there are 10+20=30 trees

So we get a maximum of 13,500 apples when there are 30 trees.

Answer 2

Let A = number of apples per tree and T= number of trees/hectare with Y= yield of apples/hectare.

Note that A=-15T+b from the narrative with A=600 with T=20, thus b=300.

Since the number of apples/hectare is being considered the funtion describing that would be Y=AT=-15T²+300T. To find extrema, Y'=-30T+300 =0 when T=10. We know Y is a maximum at T=10 because Y''=-30<0 for all T, thus T=10 is a local maximum.