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Maximums and minimums 
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#1
May210, 03:11 AM

P: 439

1. The problem statement, all variables and given/known data
y^{3}=6xyx31 , find dy/dx . Prove that the maximm value of y occurs when x^{3}=8+2sqrt(114) and the minimum value when x^{3}=82sqrt(114) 2. Relevant equations 3. The attempt at a solution I found dy/dx to be (2yx^{2})/(y^{2}2x) Then from here , dy/dx=0 for turning points and it happens when y=1/2 x^{2} Substitute this back to the original equation and find x from there using the quadratic formula [tex]x^3=\frac{16\pm\sqrt{16^24(1)(8)}}{2}[/tex] which simplifies to [tex]x^3=8\pm \sqrt{56}[/tex] First off , the x coordinate of my turning points is wrong but where is that mistake ? Also , How do i test whether this is a max or min ? Second order differentiation doesn't help .. 


#2
May210, 09:51 AM

Sci Advisor
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P: 26,157

Hi thereddevils!
(have a squareroot: √ ) The one in the book has √456 instead of √56 … perhaps someone pressed the wrong button? 


#3
May210, 10:30 AM

P: 439

I am a bit reluctant to do the second order differentiation as there are 'y's in the dy/dx equation . When i put dy/dx=0 , y= 1/2 x^2 can i substitute this into the dy/dx equation to get rid of the y ? 


#4
May210, 12:30 PM

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P: 26,157

Maximums and minimums
d^{2}y/dx^{2} = ∂/∂x(f(x,y)) + ∂/∂y(f(x,y))dy/dx the second term will be zero, and you put the x = ^{3}√(8 ± etc) and y = 1/2 x^{2} into the first term. 


#5
May210, 10:15 PM

P: 439

Can i differentiate implicitly instead of partial diff ? I tried , [tex]\frac{d^2y}{dx^2}=\frac{(y^22x)(2dy/dx2x)(2yx^2)(2y dy/dx2)}{(y^22x)^2}[/tex] when dy/dx=0 , y= 1/2 x^2 d2y/dx2=8/(x38) x= 2.49(max) or 0.80(min) 


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