# Maximums and minimums

by thereddevils
Tags: maximums, minimums
 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 $$x^3=\frac{16\pm\sqrt{16^2-4(1)(8)}}{2}$$ 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 ..
HW Helper
Thanks
P: 26,148
Hi thereddevils!

(have a square-root: √ )
 Quote by thereddevils 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.
P: 439
 Quote by tiny-tim 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 ?

HW Helper
Thanks
P: 26,148
Maximums and minimums

 Quote by thereddevils 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.
P: 439
 Quote by tiny-tim 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)
 Sci Advisor HW Helper Thanks P: 26,148 Yes, that's fine (though I haven't checked the last two lines).

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