Maximizing q in f(q)=q(Q-q) Derivation | Simplified Steps

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The discussion focuses on the derivation of the function f(q) = q(Q - q) to find the value of q that maximizes the function. The correct derivative, obtained using the product rule, is set to zero, leading to the equation Q - 2q = 0. A common mistake noted is the incorrect differentiation of the term (Q - q), where Q is treated as a constant, resulting in the derivative being -1 instead of (Q - 1). The key takeaway is the importance of correctly applying the product rule and recognizing constants during differentiation.

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I am having trouble with this derivation: f(q)=q(Q-q). We want to find the values that maximize the vaule of q so the book sets the derivative equal to 0 and gets Q-2q=0

When I try to do the derivative of f(q)=q(Q-q), I used the product rule

1(Q-q) + q(Q-1)=0
Q-q+Qq-q=0
Q+Qq-2q=0

I have an extra (Qq) I don't know how to get rid of.
 
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needhlp said:
I am having trouble with this derivation: f(q)=q(Q-q). We want to find the values that maximize the vaule of q so the book sets the derivative equal to 0 and gets Q-2q=0

When I try to do the derivative of f(q)=q(Q-q), I used the product rule

1(Q-q) + q(Q-1)=0
There's a mistake above. d/dq(Q - q) = -1, not Q - 1. Q is a constant in this function.
needhlp said:
Q-q+Qq-q=0
Q+Qq-2q=0

I have an extra (Qq) I don't know how to get rid of.
 
ah, too many q's running around. Thank you
 

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