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- Thread starter mt91
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- Thread starter
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- Jan 30, 2018

- 731

For #6, if y= Eu(E) then y'= u(E)+ Eu'(E)= 0 at a maximum. Obviously "u(x)= 0" will NOT give "maximum yield". With $u(E)= \sqrt{1- E}= (1- E)^{1/2}$, $u'= -\frac{1}{2}(1- E)^{-1/2}$ and $y'= u(E)+ Eu'(E)= (1- E)^{1/2}-\frac{1}{2}E(1- E)^{-1/2}= 0$ so $(1- E)^{1/2}= \frac{1}{2}E(1- E)^{-1/2}$ and $2(1- E)= E$, $2- 2E= E$, $2= 3E$, and $E= \frac{2}{3}$.