1. ## lagrange

as your awake how do i solve this. local max and minima from origin to curve $x^3+y^3-6xy=0$ i can't do it. answer is minima x=0=y(nb dont know working for) and max x=y=3(again don't know workin for)

2. What have you tried?

You need to solve $\bigtriangledown f(x,y)=0$

3. Originally Posted by i_zz_y_ill
as your awake how do i solve this. local max and minima from origin to curve $x^3+y^3-6xy=0$ i can't do it. answer is minima x=0=y(nb dont know working for) and max x=y=3(again don't know workin for)
first,.you need to find the critical point by
Originally Posted by pickslides
solve $\bigtriangledown f(x,y)=0$
and then use the second derivative test to determine the maxima and minima. see : Maxima and Minima of Functions of Two Variables

4. See attachment

Note to dedust and pickslides we are not looking for critical points--we want to optimize the distance fn subject to a constraint --a lagrange multiplier pblm

5. Hello, i_zz_y_ill!

I assume you know to set up Lagrnge multipliers.

Find the local max and minima from origin to curve $x^3+y^3-6xy\:=\:0$

Answer are: minimum (0, 0), maximum (3,3)

We want to minimize the distance from point $(x,y)$ to the origin.

The distance function is: . $d \:=\:\sqrt{x^2+y^2}$

For convenience, we can minimize the square of that distance: . $D \:=\:x^2+y^2$

Our function is: . $f(x,y,\lambda) \;=\;(x^2+y^2) + \lambda(x^3 + y^3 - 6xy)$

Set the partial derivatives equal to 0, and solve the system . . .

. . $\begin{array}{cccccc}f_x &=& 2x + 3\lambda x^2 - 6y &=& 0 &[1] \\

f_y &=& 2y + 3\lambda y^2 - 6x &=& 0 & [2] \\

f_{\lambda} &=& x^3 + y^3 - 6xy &=& 0 & [3] \end{array}$

$\begin{array}{ccccc}\text{From [1], we have:} & \lambda &=& \dfrac{6y-2x}{3x^2} & [4] \\ \\[-3mm]
\text{From [2], we have:} & \lambda &=& \dfrac{6x-2y}{3y^2} & [5] \end{array}$

Equate [4] and [5]: . $\frac{6y-2x}{3x^2} \:=\:\frac{6x-2y}{3y^2} \quad\Rightarrow\quad 3x^3 - 3y^3 - x^2y + xy^2 \:=\:0$

Factor: . $3(x^3-y^3) - xy(x-y) \:=\:0 \quad\Rightarrow\quad 3(x-y)(x^2+xy+y^2) - xy(x-y) \:=\:0$

Factor: . $(x-y)\bigg[3(x^2+xy+y^2) - xy\bigg] \:=\:0 \quad\Rightarrow\quad 3(x-y)(3x^2 + 2xy + 3y^2) \:=\:0$

From $x-y \:=\:0$, we have: . $\boxed{y \:=\:x}\;\;[6]$

And $3x^2 + 2xy + 3y^2 \:=\:0$ has no real factors.

Substitute [6] into [3]: . $x^3 + x^3 - 6x^2 \:=\:0 \quad\Rightarrow\quad 2x^3 - 6x^2 \:=\:0 \quad\Rightarrow\quad 2x^2(x-3) \:=\:0$

. . Hence, we have: . $x \:=\:0,3 \quad\hdots\quad y \:=\:0,3$

Therefore, the critical points are: . $(0,0),\;(3,3)$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I could be wrong, but I think this is the Folium of Descartes.

I has a leaf-shaped loop in Quadrant 1
. . . $\text{and approaches }x + y + 3\:=\:0 \text{ as an asymptote.}$

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6. Originally Posted by Calculus26
See attachment

Note to dedust and pickslides we are not looking for critical points--we want to optimize the distance fn subject to a constraint --a lagrange multiplier pblm
ah,..sorry,.i didn't read the problem carefully,.just continue what pickslides did