Answer:
The population size would be 
The yield would be    
Explanation:
 So in this problem we are going to be examining the application of a  population dynamics a fishing pond and stock fishing and objective would be to obtain the maximum sustainable yield and and the population of the fish at the obtained maximum sustainable yield,  so basically we would be applying an engineering solution to fishing 
  
     So the current  yield which is mathematically represented as 
                                
  Where dN is the change in the number of fish 
             and dt is the change in time 
So in order to obtain the solution we need to obtain the  rate of growth 
     For this we would be making use of the growth rate equation which is 
                                       ![r = \frac{[\frac{dN}{dt}] }{N[1-\frac{N}{K} ]}](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7B%5B%5Cfrac%7BdN%7D%7Bdt%7D%5D%20%7D%7BN%5B1-%5Cfrac%7BN%7D%7BK%7D%20%5D%7D)
   Where N is the population of the fish which is given as 4,000 fishes
           and  K is the carrying capacity which is given as 10,000 fishes 
              r is the growth rate 
         Substituting these values into the equation 
                               ![r = \frac{[\frac{2000}{year}] }{4000[1-\frac{4000}{10,000} ]}  =0.833](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7B%5B%5Cfrac%7B2000%7D%7Byear%7D%5D%20%7D%7B4000%5B1-%5Cfrac%7B4000%7D%7B10%2C000%7D%20%5D%7D%20%20%3D0.833)
The maximum sustainable yield would be dependent on the growth rate an the carrying capacity and this mathematically represented as 
                       
So since the maximum sustainable yield is 2082 then the the population need to be higher than 4,000 so in order to ensure a that this maximum yield the population size denoted by  would be
 would be 
                           