Weird calc question? An apple farmer can produce 600 apples...

An apple farmer can produce 600 apples from each of his apple trees if no more than 20 trees are planted. If he plants more than 20 trees the yield per tree will decrease. In fact, he figures that for each extra tree he plants, his yield per tree will decrease by 15 apples. How many trees should he plant in order to obtain the maximum number of apples?

So because we don't know what happens before x=20 I just assumed that there was a horizontal line modeled by y=600 for x≤20. Then from x is greater than 20 the graph would be modeled by a slope of -15. Then in order to find the maximum you have to figure out where that line of slope -15 crosses the x axis, because that must be the maximum amount of apples.

Using that logic I get 60 trees which is not provided on the answer sheet.
a. 450
b. 40
c. 200
d. 10
e. 30

Thanks

You have a two piece function

x in the interval [0,20] y = 600x it is good to know that at x = 20, y = 12000
x = 20 will be one endpoint
But this part of the function wont matter. We are interested in the part where x > 20

x in the interval (20, 60) = 600x - 15x(x - 20)
WHY? 600 apples per tree planeted = 600x
15 apples less per tree (15x) for each tree over 20 that is planted (x - 20)

600x + 300x - 15x^2 rearrange into descending order and combine like terms
- 15x^2 + 900x This is the function for the yield per tree when x > 20

d/dx [- 15x^2 + 900x] = - 30x + 900

- 30x + 900 = 0 --> x = 30

I take it you derived y - 600 = -15(x - 20), or y = -15x + 900, where x is the number of trees planted, and y is the yield per tree. Then the total number A of apples produced is the number of trees planted times the yield per tree, so you want to maximize A = x*y, which is -15x² + 900x. You will readily find that dA/dx = 0 ⇒ x = 30.