# Math Help - Find max and min

1. ## Find max and min

This problem I think it can be soved with Lagrange multipliers but I tried to solve this with out them.
Find the maximum and minimum values of the variable x on the Equation:

$4y^{2}-2xy+x^{2}=3$

I dont know if Im having all the solutions when I solve for y:
$(2y - \frac{1}{2}x)^{2}+\frac{3}{4}x^2=3$

$y = \dfrac{\sqrt{3 - \frac{3}{4}x^{2}} + \frac{1}{2}x}{2}$

$\sqrt{3 - \frac{3}{4}x^{2}} > 0$

Solving it I have:

$|x| > 2$

2. Originally Posted by kezman
This problem I think it can be soved with Lagrange multipliers but I tried to solve this with out them.
Find the maximum and minimum values of the variable x on the Equation:

$4y^{2}-2xy+x^{2}=3$

I dont know if Im having all the solutions when I solve for y:
$(2y - \frac{1}{2}x)^{2}+\frac{3}{4}x^2=3$

$y = \dfrac{\sqrt{3 - \frac{3}{4}x^{2}} + \frac{1}{2}x}{2}$

$\sqrt{3 - \frac{3}{4}x^{2}} > 0$

Solving it I have:

$|x| > 2$
The problem is that there are 2 functions.
When you take the square root (asuming it is positive) you get two different functions. One for positive and one for negative.

3. Hello, kezman!

If I read the problem correctly, your approach is wrong . . .

Find the maximum and minimum values of x on the equation:

. . $4y^{2}-2xy+x^2\:=\:3$

If we're supposed to maximize/minimize $x$, we must solve for $x.$

We have the quadratic: . $x^2 - 2yx + 4y^2 - 3 \:=\:0$

Quadratic Formula: . $x \;=\;\frac{2y \pm\sqrt{4y^2 - 4(4y^2-3)}}{2}$

. . and we have: . $x \;=\;y \pm \sqrt{3-3y^2}$

And that is the function we must maximize/minimize.

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

With Lagrange multipliers, we want to maximize $x$
. . with the constraint: $4y^2 - 2xy + x^2 \:= \:3$

So we have: . $F(x,y,\lambda) \:=\:x + \lambda(4y^2 - 2xy + x^2 - 3)$