Raymond B. answered 04/14/23
Math, microeconomics or criminal justice
O= (60+t)(500-5t)= 3000+500t-300t- 5t^2
0ranges=3,000+200t-5t^2 where t=number of additional trees
he should add 20 trees for total 80 trees to get 5,000 Oranges =max possible
for 50 original trees and 400 oranges per tree originally, and 4 oranges less per extra tree:
O=(50+t)(400-4t)= 2000+200t-4t^2
then add 25 trees to get 75(300) = 22,500 Oranges = max
more generally
O=(T+t)(N-nt) where T=original trees, t=extra trees,N=original number of oranges per original Trees, n=reduction of oranges per tree
O(t)= TN+(N-n)t - nt^2
O'(t) = N-n -2nt = 0
t=(N-n)/2n = N/2n -1/2 = optimal number of extra trees
max Oranges = O(N/2n-1/2) = TN+(N-n)(N/2n-1/2) -n(N/2n-1/2)^2