Turning points

[IAIN GLENNY]

Stuck on this due to mixed terms ....any help much appreciated ....gone back to uni after raising kids and my maths is not what it was

"locate turning points and determine if they are max/min..

F(x,y)= 2x^3 + 6xy^2 - 3y^3 -150x

 

[stapel]

"locate turning points and determine if they are max/min..

F(x,y)= 2x^3 + 6xy^2 - 3y^3 -150x

What definition have they given you for "turning points"? What rules or methods have they given you for determining max/min points? How have you applied this information? Where are you getting stuck?

Please be complete. Thank you!

 

[IAIN GLENNY]

Its a calculus paper ... Cant figure out how to seperate terms so i can derive where gradient = 0 not entirely sure if f (x,y) is function of x in respect to y or the function of x and y ....? No further info

 

[HallsofIvy]

Are you refusing to answer stapel's questions or do you really not know the definitions of any of the words you are using?

The function given f(x, y) is a function of both x and y. y is NOT a function of x and x is NOT a function of y. Now, again, what is the definition of a "turning point" of a function of two variables? What is the gradient of f?

 

[IAIN GLENNY]

GRADIENT IS DY / DX ...TURNING POINT IS WHERE GRADIENT CHANGES FROM NEGATIVE TO POSITIVE . NOT ENTIRELY SURE WHERE THIS IS GOING .

fOR EXAMPLE PREVIOUS PAPER QUESTION . WAS Y=3X^2 - 6X THEREFOR DY/DX = 6X-6 SO WHEN X=1 DY/DX=6-6=0 WHICH IS MIN TURNING POINT ... NO WORRIES THERE HOWEVER WOULD APPRECIATE SOME HELP UNDERSTANDING INITIAL QUESTION....


THANKYOU

 

[stapel]

GRADIENT IS DY / DX

Only if you are dealing with y as a function of x. But you're not; you're dealing with F as a function of x and y. So, as pointed out earlier, dy/dx is NOT the derivative (or "slope at a point" or "gradient").

 

[HallsofIvy]

GRADIENT IS DY / DX ...TURNING POINT IS WHERE GRADIENT CHANGES FROM NEGATIVE TO POSITIVE . NOT ENTIRELY SURE WHERE THIS IS GOING .

No, it isn't. You are still trying to make y a function of x. This is a multi-variable calculus problem, not a "Calculus I", single variable problem.

fOR EXAMPLE PREVIOUS PAPER QUESTION . WAS Y=3X^2 - 6X THEREFOR DY/DX = 6X-6 SO WHEN X=1 DY/DX=6-6=0 WHICH IS MIN TURNING POINT ... NO WORRIES THERE HOWEVER WOULD APPRECIATE SOME HELP UNDERSTANDING INITIAL QUESTION....


THANKYOU

Given F(x,y), a function of two variables, its "gradient" is the vector, $\displaystyle \frac{\partial f}{\partial x}\vec{i}+ \frac{\partial F}{\partial y}\vec{j}$.

It always points in the direction of fastest increase of the function and its length is the rate of increase in that direction. At a point at which the function is a maximum, there is NO "direction of fastest increase" because the function does not increase in any direction. The gradient vector must be 0.