Answer:
The population that gives the maximum sustainable yield is 45000 swordfishes.
The maximum sustainable yield is 202500 swordfishes.
Step-by-step explanation:
Let be [tex]f(p) = -0.01\cdot p^{2}+9\cdot p[/tex], the maximum sustainable yield can be found by using first and second derivatives of the given function: (First and Second Derivative Tests)
First Derivative Test
[tex]f'(p) = -0.02\cdot p +9[/tex]
Let equalize the resulting expression to zero and solve afterwards:
[tex]-0.02\cdot p + 9 = 0[/tex]
[tex]p = 450[/tex]
Second Derivative Test
[tex]f''(p) = -0.02[/tex]
This means that result on previous part leads to an absolute maximum.
The population that gives the maximum sustainable yield is 45000 swordfishes.
The maximum sustainable yield is:
[tex]f(450) = -0.01\cdot (450)^{2}+9\cdot (450)[/tex]
[tex]f(450) =2025[/tex]
The maximum sustainable yield is 202500 swordfishes.