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Calculus/Relative max/min

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Expert: Alon Mandes - 11/3/2008


Question
I was given y=x(sqrt(3-x)), and I need to find the relative max and min.  I found the first derivative to be (3-x)^(1/2)-((x)/2(3-x)^(-1/2)), but since there is a square root in the bottom, I don't know what to do.  Please Help!

Answer
Hey Molly, let's review some facts about derivating square roots :
1. [√x]'=1/2√x
2. [√(3-x)]'=-1/2√(3-x). "The - sign because (3-x)'=-1.
3. [fg]'=f'g+g'f , SO
[x√(3-x)]'=x'√(3-x)+x[√(3-x)]'=√(3-x)-x/2√(3-x).
Let's set this form equal zero :
√(3-x)-x/2√(3-x)=0 -> √(3-x)=x/2√(3-x) -> 2(3-x)=x -> 6-2x=x ->
x=2. Now let's chick if this result is max or min :
For that we have to calculate the 2nd derivative :
y''=[√(3-x)-x/2√(3-x)]'=1/2√(3-x) -(1/2)/2√(3-x)-(x/2)*(3-x)^(3/2)
This is true because 1/√(3-x)=(3-x)^(-1/2) &
[(3-x)^(1/2)]'=(-1)*(-1/2)(3-x)^(3/2). So,
y''=1/2√(3-x)-(1/2)/2√(3-x)-(x/2)*(3-x)^(3/2)
y''(2)=1/2-(1/4)-1=-3/4 <0 , so min

Alon.

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Alon Mandes

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Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems. Kind of question I can't answer : Economics,Combinatorics,infinite series & convergence ,Statistics & Probabilities .

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1. I'm a team member of mathnerds (math site for answering questions) 2. I'm a team member in the Student's Union of the Technion, helping students who have problems in mathematics. 3. 2 years of experience as a math teacher in college. 4. I give free homework help for high school students in Mathematics & Physics. 5. I teach part time in collage the subjects : "Digital Signal Processing" , "Random Signals & Noise" , "Complex Functions".

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Hi-Tech company : GSM4VOIP ; job possition : Algorythm developer.

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M.A in Mathematics & Bs.c in Electronics.

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