Optimizing y3=6xy-x3-1: Max and Min Values | Calculating dy/dx

[thereddevils]

Homework Statement

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)

Homework Equations

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

x^3=\frac{16\pm\sqrt{16^2-4(1)(8)}}{2} which simplifies to

x^3=8\pm \sqrt{56}

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 ..

 

[tiny-tim]

Hi thereddevils!

(have a square-root: √ )

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

x^3=\frac{16\pm\sqrt{16^2-4(1)(8)}}{2} which simplifies to

x^3=8\pm \sqrt{56}

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]

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]

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]

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 ,

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

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

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

x= 2.49(max) or 0.80(min)

 

[tiny-tim]

Yes, that's fine (though I haven't checked the last two lines).