a certain species of fish will grow from x million to x(15-x) million each year. In order to sustain a steady catch each year, a limit of x(15-x)-x million fish are to be caught, leaving x million fish to reproduce each year. What is the number of fish which should be left to reproduce each year so that the maximum catch may be sustained year to year?
A) 5 million
B) 7 million
C) 7.5 million
D)10 million
E)15 million
Can you please explain to me how to do this? thank you!
2007-03-11 09:54:23 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
why do we take derivative of y = x(15-x)?
2007-03-11 10:01:36 · update #1
Since the others have the logic covered, I'll answer this:
"why do we take derivative of y = x(15-x)?"
Because the first derivative is a mark of growth / recession. Or, in this case, the number of fish which are created / born each year.
2007-03-11 10:48:01 · answer #1 · answered by Dan Lobos 2 · 0⤊ 0⤋
y = x(15-x)-x
dydx = 15 -2x -1 =14-2x
14 - 2x =0 to get max
2x = 14
x=7 million
2007-03-11 10:34:27 · answer #2 · answered by hbj 2 · 0⤊ 0⤋
y = x(15-x)-x
dydx = 15 -2x -1 =14-2x
14 - 2x =0 to get max
2x = 14
x=7 million
2007-03-11 10:12:52 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
y = x(15-x)
y' = 15-x -x = 0
x = 7.5 millions
2007-03-11 09:59:40 · answer #4 · answered by sahsjing 7 · 0⤊ 1⤋