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## Homework Statement

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

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- Thread starter Liang Wei
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- #1

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

- #2

PhysicoRaj

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

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

- #3

PhysicoRaj

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What is y defined to be?

- #4

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-x^2)

- #5

PhysicoRaj

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

- #6

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Please post the full question.

- #7

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

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- #9

PhysicoRaj

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Okay,

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

and

##x^3=8\pm2\sqrt14##

right?

Now how have you decided to proceed?

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

and

##x^3=8\pm2\sqrt14##

right?

Now how have you decided to proceed?

- #10

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Yup correct

- #11

PhysicoRaj

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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?

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- #13

PhysicoRaj

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Right.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

Ok that will fetch you the maximum and minimum of y.then I substitute x=2+2(14)^(1/2) and x=2-2(14)^(1/2) to get y values right?

I'm not sure if I understood this one.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.

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

- #14

PhysicoRaj

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- #15

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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 ? @@

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- #17

PhysicoRaj

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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?but as I sub it back into y^3

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- #19

PhysicoRaj

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- #20

PhysicoRaj

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If typing in was difficult, a snapshot of your work will do.

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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?@@

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- #23

PhysicoRaj

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

- #24

PhysicoRaj

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