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OpenStudy (anonymous):
OpenStudy (anonymous):
so i should probably take the limit?
ganeshie8 (ganeshie8):
find the derivative,
set it equal to 0 and find ur critical points ?
ganeshie8 (ganeshie8):
i presume u knw optimization as i saw u doing integrals in ur past posts
OpenStudy (anonymous):
Im still working the numbers, I'm missing something for sure..
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ganeshie8 (ganeshie8):
did u find the derivative ?
ganeshie8 (ganeshie8):
\(Y(N) = \dfrac{N}{1+N^2}\)
\(Y'(N) = ?\)
OpenStudy (anonymous):
yeah i got (N)(-1)(1+N)^(-2) + (1+N)^-1
ganeshie8 (ganeshie8):
one sec leme check
OpenStudy (anonymous):
uhg i know what i did, i took y'(n) or (1+N)^-1 instead of -2 power
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OpenStudy (anonymous):
of*
ganeshie8 (ganeshie8):
hahah try again
ganeshie8 (ganeshie8):
use quotient/product rule whichever you're friendly wid
OpenStudy (anonymous):
so i got N = 1
OpenStudy (anonymous):
after setting my new derivative to zero
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ganeshie8 (ganeshie8):
Excellent !
ganeshie8 (ganeshie8):
so, N = 1 is your only critical value
OpenStudy (anonymous):
so the slope of my graph is zero at x = 1 therefore that is a min or a max
OpenStudy (anonymous):
so i need to find if it is increasing or decreasing on either side right?
ganeshie8 (ganeshie8):
Yep !
you got it !! it better give a max value as the graph is decreasing (intuitively)
use second derivative to figure out whether its min/max/neither
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ganeshie8 (ganeshie8):
find second derivative,
evaluate it at 1
ganeshie8 (ganeshie8):
f''(1) is negative => its a max
OpenStudy (anonymous):
it is so Nitrogen level 1 maximizes the yield
OpenStudy (anonymous):
thank you :)
ganeshie8 (ganeshie8):
Correct ! graph it and see for double confirmation :)
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OpenStudy (anonymous):
yeah ill graph it now, i was trying to do it without, and we did :D
OpenStudy (anonymous):
WFA sometimes prevents me from learning
ganeshie8 (ganeshie8):
ganeshie8 (ganeshie8):
lol true, WFA's job is to check ur answers... i bet you're wise enough to cheat urself :P