Maximum and minimum values

[Liang Wei]

Homework Statement

Show that the maximum and minimum values of y occurs when x^3=8+- 2(14)^1/2

Homework Equations

The Attempt at a Solution

 

[PhysicoRaj]

What have you attempted at? Do you know the method to solve this?

Plus check this: x^3=8+- 2(14)^1/2

is it : $x^3 = 8+(-2)(14)^{1/2}$ ??

 

[PhysicoRaj]

What is y defined to be?

 

[Liang Wei]

It should be +on the top and minus on the bottom,not sure how to write it and I have been using the sign test to test it,however the sign test should be for y values,but only x values are given by the question,plus do I sub the x values into the y^3 equation to find y or should I use dy/dx =(2y-x^2)/(y^2-2x)=0 then sub x values to find y?
-x^2)

 

[PhysicoRaj]

You should have definition for y, right? Else what would you differentiate?

 

[Quesadilla]

Please post the full question.

 

[Liang Wei]

Y^3=6xy-x^3-1

 

[Liang Wei]

I think I have posted the attachment now,sorry I am a first timer here

 

[PhysicoRaj]

Okay,

$y^3=6xy-x^3-1$

and

$x^3=8\pm2\sqrt14$

right?

Now how have you decided to proceed?

 

[Liang Wei]

Yup correct

 

[PhysicoRaj]

I think I have posted the attachment now,sorry I am a first timer here

Well, that completely makes sense now. What have you attempted?

 

[Liang Wei]

As I know at stationary point,dy/dx=0 so from dy/dx=(2y-x^2)/(y^2-2x),from there I can find y=(x^2)/2,then I substitute x=2+2(14)^(1/6) and x=2-2(14)^(1/6) to get y values right? Then I reciprocate the dy/dx =dx/dy and then I do the sign test? But I got all + signs which means I fail to prove y has maximum and minimum values.

 

[PhysicoRaj]

As I know at stationary point,dy/dx=0 so from dy/dx=(2y-x^2)/(y^2-2x),from there I can find y=(x^2)/2

Right.

then I substitute x=2+2(14)^(1/2) and x=2-2(14)^(1/2) to get y values right?

Ok that will fetch you the maximum and minimum of y.

Then I reciprocate the dy/dx =dx/dy and then I do the sign test? But I got all + signs which means I fail to prove y has maximum and minimum values.

I'm not sure if I understood this one.

But as soon as you get the $y=x^2/2$, can you use it and the 'given' to solve for values of x?

 

[PhysicoRaj]

Note: the question is to show that the maximum and minimum of y occurs at x=so and so. Not to find the maximum and minimum values of y itself. So, you have to arrive at the x values, rather than substituting them.

 

[Liang Wei]

Right.


Ok that will fetch you the maximum and minimum of y.


I'm not sure if I understood this one.

But as soon as you get the $y=x^2/2$, can you use it and the 'given' to solve for values of x?

I haven't attempt which I didn't thought of it that but as I sub it back into y^3, I get x=2 and y=2,so now I have 3 values of x and I can find 3 values of y,and then I need to sub into the second derivative test or using a sign test but If I do a sign test there,there are two variables,how to do it ? @@

 

[Liang Wei]

Oo then by using the 3 values of x,I do a second derivative test and sub the values in order to prove that the x values are maximum and minimum values and we can show that maximum and minimum value of y occurs at those x right? And can I do a sign test to test whether y is decreasing or increasing?

 

[PhysicoRaj]

but as I sub it back into y^3

You mean you substituted y=x^2/2 in the original y equation. What did it yield? Which are the three x's and y's?

 

[Liang Wei]

Yup I sub back into the original y equation and get another value of x and y ,however I just noticed that I don't actually need it right? Because I can do a 2nd derivative and put the previous x and y values into it and find whether it is maximum or minimum values,but again may I use a sign test for an implicit differentiation equation to find maximum and values oo?

 

[PhysicoRaj]

No, if you can find the points of maxima and minima without using the maximum and minimum values, it's alright then. But if you require them to solve for the points of inflection, you will have to find the max. and min. values.

 

[PhysicoRaj]

Can you post what you actually worked out? Because it's difficult to imagine what you have done or find the point where you are lacking.
If typing in was difficult, a snapshot of your work will do.

 

[Liang Wei]

I just tried the 2nd derivative

I have used the calculator to change the x^3 values into 15.84 and -5.84,but when I substitute the x values and y values (how I find y values,I get it from y=(x^2)/2,substituting the x values as stated above) into my second derivative but in the end I got all positive value which means only local minimum values only. So I did it wrongly?@@

 

[Liang Wei]

This is another method that I attempted to do but it is wrong

 

[PhysicoRaj]

I have used the calculator to change the x^3 values into 15.84 and -5.84

How did you get this? Probably by the previous x^3 equation which they have given?

but when I substitute the x values and y values into my second derivative but in the end I got all positive value which means only local minimum values only.

The implication is right, but I don't know if all are minima or maxima or both, because I haven't worked it out. [STRIKE]If all are minima, that doesn't have anything to do with the answer they expect.[/STRIKE] Okay I saw what the question asks for.

 

[PhysicoRaj]

I think they expect you to arrive at $x^3=8+-2(14)^{1/2}$ rather than assume it at the first. This is what is going wrong.

 

[Liang Wei]

Yeah,the x values are from the x^3 that they provide,when you substitute those values,you will find that there are many zero values so it is able to get the d2y/dx2= +value which is greater than zero so both values of x are local minimums

 

[Liang Wei]

Oooo

 

[PhysicoRaj]

Yeah,the x values are from the x^3 that they provide,when you substitute those values,you will find that there are many zero values so it is able to get the d2y/dx2= +value which is greater than zero so both values of x are local minimums

Okay I see..
Go step by step: You found first derivative and equated it to zero. Implies y=x^2/2. Next step is to use this and solve for 'x'. See if you can do this by substituting in the givens.

 

[Liang Wei]

I think they expect you to arrive at $x^3=8+-2(14)^{1/2}$ rather than assume it at the first. This is what is going wrong.

I don't think I know how do I get x^3=8+-2(14)^{1/2} from the equations provided

 

[PhysicoRaj]

I don't think I know how do I get x^3=8+-2(14)^{1/2} from the equations provided

I know, it is quite difficult. What you have done is right, but there are several doubts like:
>You used a calculator to approximate the given x value ( I think this is unfair).
>Both x values are yielding minima (right?)

 

[Liang Wei]

Yup both x values are yielding minima,if I don't use the calculator,it would be really difficult to sub it into the 2nd derivative. It has took me several hours but I am still unable to work it out.

 

[PhysicoRaj]

In the first line of the image in post #21, while finding the second derivative, you have written -2x instead of y^2-2x

 

[Liang Wei]

Erm I have found my mistakes and able to show it but an additional question is can I use the sign test to find the maximum and minimum values from this kind of equation?sorry to bother you and thanks.

 

[PhysicoRaj]

can I use the sign test to find the maximum and minimum values from this kind of equation?

You mean the second derivative test? It says only if the point is maxima or minima. It doesn't give you the value.

 

[Liang Wei]

Sign test is a test which you use a value which lies inside the interval of a variable,example x then substitute it into dy/dx to find the sign and we can find the maxima and minima.but anyways I manage to prove it by using 2nd derivative but not sure correct or wrong.

 

[PhysicoRaj]

Sign test is a test which you use a value which lies inside the interval of a variable,example x then substitute it into dy/dx to find the sign and we can find the maxima and minima.

Oh that one, for that you need to know the interval, and it requires you to solve for x.

but anyways I manage to prove it by using 2nd derivative but not sure correct or wrong.

Correct as long as you don't assume the values of x with a calculator and as long as y has both maxima and minima.