The fish population in a pond with carrying capacity 1200 is modeled by the following logistic equation where N(t) denotes the number of fish at time t in years. (dN)/(dt) = (0.4)/(1200)N(1200-N)
Starting at the time at which the number of fish reached 150, the owner of the pond removed (harvested) fish at a constant rate of 50 fish per year. We take t = 0 to be the time at which the owner started to harvest fish from the pond.

A) Modify the differential equation to model the population of fish from the time it reached 150.
dN/dt =

B) How far below the original carrying capacity will the number of fish be in the long run? (Give your answer correct to the nearest whole fish.)
Number of fish=

9514 1404 393

  (dN)/(dt) = (0.4)/(1200)N(1200-N) -50

  142 fish

Step-by-step explanation:

A) The differential equation is modified by adding a -50 fish per year constant term:

  (dN)/(dt) = (0.4)/(1200)N(1200-N) -50

__

B) The steady-state value of the fish population will be when N reaches the value that makes dN/dt = 0.

  (0.4/1200)(N)(1200-N) -50 = 0

  N(N-1200) = -(50)(1200)/0.4) . . . . rewrite so N^2 has a positive coefficient

  N^2 -1200N + 600^2 = -150,000 +600^2 . . . . complete the square

  (N -600)^2 = 210,000 . . . . . simplify

  N = 600 + √210,000 ≈ 1058

This steady-state number of fish is ...

  1200 - 1058 = 142 . . . . below the original carrying capacity

$20.01

step-by-step explanation:

the sale price is (1.00 - 0.77)($87), or 0.23($87), or $20.01.

y= 2x+7

changed to y= 5x +7

as slope is increased from 2 to 5

therefore it would be less steeper

so, option d) is correct