1. ## Exponential Differentiation

In studying salmon populations, a model often used is the Ricker equation which relates the size of a fish population this year,
x to the expected size next year y. (Note that these populations do not change continuously, since all the parents die before the eggs are hatched.) The Ricker equation is
y = axe^(-bx) where a,b> 0
Find the value of the current population which maximizes the salmon population next year according to this model.

---> I differentiated it this far:
dy/dx = axe(-bx)(-b) + e^(-bx)(a)
dy/dx = -axe^(-bx)b + ae^(-bx)
dy/dx = 0 = ae^(-bx) (-xb + 1)

in the question we are given that a,b>0 Could anyone please tell me how to further solve this question??

2. Originally Posted by ninja
in studying salmon populations, a model often used is the ricker equation which relates the size of a fish population this year,
x to the expected size next year y. (note that these populations do not change continuously, since all the parents die before the eggs are hatched.) the ricker equation is
y = axe^(-bx) where a,b> 0
find the value of the current population which maximizes the salmon population next year according to this model.

---> i differentiated it this far:
dy/dx = axe^(-bx)(-b) + e^(-bx)(a)
dy/dx = -abxe^(-bx) + ae^(-bx)
dy/dx = 0 = ae^(-bx) (-bx + 1)

$\displaystyle \textcolor{red}{x = \frac{1}{b}}$

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3. Originally Posted by skeeter

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i was just doing something wrong in the second derivative (to prove max) due to which it was turning out be a minimum. thanks anyways for spending your time on it.