How to find the maximum and minimum value by using Lagrange Multipliers

Question

Find the maximum and minimum values of $f(x,y)=xy$

subject to $\frac{y^2}{2}+\frac{x^2}{4}=1$

Write the Lagrange function in the form $f(x,y) = \lambda g(x,y)$

So far I've got $x=\sqrt2$ or $2\sqrt2$ and $y=1,-1$ or $y=2,-2$

I'm kinda stuck from here and not sure what to do next, can anyone please help me with the question?

Answer 1

$\frac{x^2}{4} + \frac{y^2}{2} = 1 \,$ or $x^2 + 2y^2 = 4 \implies (x - \sqrt2 \, y)^2 + 2 \sqrt2 xy = 4$

$\implies xy = \sqrt2 - \frac{(x -\sqrt2 \, y)^2}{2 \sqrt2} $. The min value of $(x -\sqrt2 \, y)^2$ is zero at $x = \sqrt 2y,$ (where $xy \ne 0)$

That gives max value of $xy = \sqrt2$.

Similarly $(x + \sqrt2 \, y)^2 - 2 \sqrt2 xy = 4 \implies xy = -\sqrt2 + \frac{(x -\sqrt2 \, y)^2}{2 \sqrt2} $

And that gives min value of $xy = -\sqrt2$.

But as the question is specifically asking to use Lagrange method,

Given constraint, $\frac{x^2}{4} + \frac{y^2}{2} = 1$ ..(i)

using Lagrange multiplier, $xy = \lambda (\frac{x^2}{4} + \frac{y^2}{2} - 1)$

Taking partial derivatives, you get -

$y = \frac{ \lambda x}{2}, x = \lambda y$

So you get $x^2 = 2y^2\, (xy \ne 0)$ and substituting in (i), $y = \pm 1, x = \pm \sqrt 2$

So maximum value of $xy = \sqrt2$, minimum value = $-\sqrt2$.

Answer 2

Plug the $ (x,y)$ pairs into the function and see which one gives the maximum. It's trial and error at this point.

A little more commentary:

The Lagrange multiplier method gives the condition for an $(x,y)$ point to be maximum or minimum. Once you got this set of points, you have to search among the points to see which one is the one which is helpful in the objective you want to do.

Actually, there is a shortcut, you can use the second derivatives (similar to the single variable) to classify maximum and minima. See more about it here

Answer 3

Let $f(x,y)=xy+\lambda\left(\frac{x^2}{4}+\frac{y^2}{2}-1\right).$

Thus, for the extreme point we obtain: $$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0,$$ which gives the following system: $$y+\frac{1}{2}\lambda x=x+\lambda y=0.$$ Now, if $xy=0$ so $x=y=0,$ which is a contradiction with our condition.

Thus, $xy\neq0$ and we obtain: $$\frac{2y}{x}=\frac{x}{y}$$ or $$x^2=2y^2,$$ which gives needed points and extreme values.

I think, it's better to use the following.

By AM-GM $$1=\frac{x^2}{4}+\frac{y^2}{2}\geq2\sqrt{\frac{x^2}{4}\cdot\frac{y^2}{2}}=\frac{|xy|}{\sqrt2},$$ which gives $$|xy|\leq\sqrt2$$ or $$-\sqrt2\leq xy\leq\sqrt2.$$ The equality occurs for $$\frac{x^2}{4}=\frac{y^2}{2}$$ or $$x^2=2y^2,$$ which says that we got a maximal value and the minimal value.

Answer 4

With the binding constant $\lambda $ in the Lagrangian

$$ f(x,y) + \lambda g(x,y) $$

$$xy+\lambda\left(\frac{x^2}{4}+\frac{y^2}{2}-1\right)$$

we have successive partial derivatives

$$ -\lambda =\dfrac{y}{x}= \dfrac{\dfrac{\partial f(x,y)}{\partial x}}{ {\dfrac{\partial f(x,y)}{\partial y}} }=\dfrac{\dfrac{\partial g(x,y)}{\partial x}}{ {\dfrac{\partial g(x,y)}{\partial y}} }=\dfrac{x/2}{y} $$

gives on cross multiplication the solution of the pair of straight lines

$$ \dfrac{x^2}{2}= y^2$$

The question should perhaps be.. find extreme value of $xy$ etc., it seeks to find maximum yellow area $4xy$ in the given ellipse. Any other rectangle has area less than what is given by this solution.