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Maximums and minimums

by thereddevils
Tags: maximums, minimums
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thereddevils
#1
May2-10, 03:11 AM
P: 439
1. The problem statement, all variables and given/known data

y3=6xy-x3-1 , find dy/dx . Prove that the maximm value of y occurs when
x3=8+2sqrt(114) and the minimum value when x3=8-2sqrt(114)

2. Relevant equations



3. The attempt at a solution

I found dy/dx to be (2y-x2)/(y2-2x)

Then from here , dy/dx=0 for turning points and it happens when y=1/2 x2

Substitute this back to the original equation and find x from there using the quadratic formula

[tex]x^3=\frac{16\pm\sqrt{16^2-4(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 ..
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tiny-tim
#2
May2-10, 09:51 AM
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Hi thereddevils!

(have a square-root: √ )
Quote Quote by thereddevils View Post
Substitute this back to the original equation and find x from there using the quadratic formula

[tex]x^3=\frac{16\pm\sqrt{16^2-4(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 ?
Your solution looks correct to me.

The one in the book has √456 instead of √56 perhaps someone pressed the wrong button?
Also , How do i test whether this is a max or min ?

Second order differentiation doesn't help ..
It should do just differentiate your equation for dy/dx wrt x, and remember you can put dy/dx = 0.
thereddevils
#3
May2-10, 10:30 AM
P: 439
Quote Quote by tiny-tim View Post
Hi thereddevils!

(have a square-root: √ )


Your solution looks correct to me.

The one in the book has √456 instead of √56 perhaps someone pressed the wrong button?


It should do just differentiate your equation for dy/dx wrt x, and remember you can put dy/dx = 0.
thanks Tiny ! I am still a little confused with the max and mins .

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 ?

tiny-tim
#4
May2-10, 12:30 PM
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Maximums and minimums

Quote Quote by thereddevils View Post
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 ?
Yes, if you already have dy/dx = f(x,y), then d/dx it to get

d2y/dx2 = ∂/∂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 x2 into the first term.
thereddevils
#5
May2-10, 10:15 PM
P: 439
Quote Quote by tiny-tim View Post
Yes, if you already have dy/dx = f(x,y), then d/dx it to get

d2y/dx2 = ∂/∂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 x2 into the first term.
erm , that looks to be partial differentiation and i am not so familiar with that but i would like to learn that

Can i differentiate implicitly instead of partial diff ?

I tried ,

[tex]\frac{d^2y}{dx^2}=\frac{(y^2-2x)(2dy/dx-2x)-(2y-x^2)(2y dy/dx-2)}{(y^2-2x)^2}[/tex]

when dy/dx=0 , y= 1/2 x^2

d2y/dx2=-8/(x3-8)

x= 2.49(max) or 0.80(min)
tiny-tim
#6
May3-10, 02:18 AM
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Yes, that's fine (though I haven't checked the last two lines).


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