- #1

arhzz

- 241

- 46

- Homework Statement
- Find the extreme values

- Relevant Equations
- -

Hello!

I am suspossed to find the extreme values of this function

$$f(x,y) = x^3 +3x^2y+3y^2-6y $$

First thing I did is I found all the partial derivations and set them to 0.Since that has to be true for the extreme value to even exist.

The derivations look like this

$$ f_x = 3x^2+6xy $$

$$ f_y = 3x^2+6y-6 $$

Now I solved this systems of equation by finding an expression for y out of the second equation ## y = \frac{-x^2}{2} +1 ## and I plug that in the first equation I get this after some basic algebra

$$ 3x(x-x^2+2) = 0 $$ Now this is going to be 0 when x is equals to 0 or when this in the bracket equals zero.So one zero is already found and its 0,so x1 = 0 and the other 2 should be x2 = 2 and x3 = -1. Now I've pluged each x back into the expression for y to get all the y values.Should be

y1 =1, y2 = -1, y3 = 1/2 and now plugged both x and y in the given function to get the z component.I get this

z1 = -3, z2 = 5 and z3 = 7/4 Now here is the kicker.A buddy of mine calculated this example as well and we get the same values for x,but he assigned after the quadratic equation x2 to be = -1 and x3 to be 2 and his z component is totally diffrent than mine.Now I am not sure if he or I made somwhere an error in the calculation,or when we get to the end (My next step would be to do the Hesse Matrix) we would get the same.But this raises a question,because we checked our calculations and it seems right but the results are just diffrent.Does it matter how we assing the values of x,and y for that matter?

Thanks!

I am suspossed to find the extreme values of this function

$$f(x,y) = x^3 +3x^2y+3y^2-6y $$

First thing I did is I found all the partial derivations and set them to 0.Since that has to be true for the extreme value to even exist.

The derivations look like this

$$ f_x = 3x^2+6xy $$

$$ f_y = 3x^2+6y-6 $$

Now I solved this systems of equation by finding an expression for y out of the second equation ## y = \frac{-x^2}{2} +1 ## and I plug that in the first equation I get this after some basic algebra

$$ 3x(x-x^2+2) = 0 $$ Now this is going to be 0 when x is equals to 0 or when this in the bracket equals zero.So one zero is already found and its 0,so x1 = 0 and the other 2 should be x2 = 2 and x3 = -1. Now I've pluged each x back into the expression for y to get all the y values.Should be

y1 =1, y2 = -1, y3 = 1/2 and now plugged both x and y in the given function to get the z component.I get this

z1 = -3, z2 = 5 and z3 = 7/4 Now here is the kicker.A buddy of mine calculated this example as well and we get the same values for x,but he assigned after the quadratic equation x2 to be = -1 and x3 to be 2 and his z component is totally diffrent than mine.Now I am not sure if he or I made somwhere an error in the calculation,or when we get to the end (My next step would be to do the Hesse Matrix) we would get the same.But this raises a question,because we checked our calculations and it seems right but the results are just diffrent.Does it matter how we assing the values of x,and y for that matter?

Thanks!

Last edited: