# Using Lagrange multipliers for restricted extrema

Consider the function $f(x,y) = x^2 + xy + y^2$ defined on the unit disc $D = \{(x,y) \mid x^2 + y^2 \leq 1\}$.

I can not simplify the equations to the point where I find a constant for the lagrange multiplier and thus can't find the points of the extrema. I used the method that we can create a new function $L$ with the variables $x$, $y$ and $\lambda$ where $\lambda$ is the Lagrange multiplier. Then found the partials of $L$ with respect to each variable. This is where I am stuck because I can not simplify enough to find $x$ and $y$. Are there any tricks for this type of question?

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Are you finding all extrema (mins and maxes) or just mins...? –  icurays1 Dec 2 '12 at 22:46
I am finding both, but I actually solved the question. Thanks for commenting though –  tamefoxes Dec 2 '12 at 22:47

The function is continuous and differentiable, so its maximum value over a region will be critical points or at the boundary.

To find critical points, we find $f_x$ and $f_y$ and set them both equal to $0$: $$f_x = 2x+y = 0$$ $$f_y = 2y+x = 0$$ $$2(-2x)+x = 0$$ $$x = 0, y = 0$$

So the origin is a critical point. Using the second derivative test in two dimensions: $$f_{xx} = 2$$ $$f_{xy} = f_{yx} = 1$$ $$f_{yy} = 2$$ Because $f_{xx}*f_{yy} - f_{xy}^2 = 3 > 0$ and $f_{xx}*f_{yy} > 0$ (so its a min, not a max), $(0,0)$ is a minimum, so it cannot possible be the greatest value.

Now we have to test the boundary, as the greatest value has to be there. Our new constraint is $g(x,y) = x^2+y^2 = 1$, so $\nabla g = <2x, 2y>$

Using Lagrange multipliers, we know $\nabla f = \lambda \nabla g$: $$2x+y = \lambda 2x$$ $$2y +x = \lambda 2y$$

$${2x+y \over 2x} = {2y+x \over 2y}$$ $$4xy+2y^2 = 4xy + 2x^2$$ $$x^2 = y^2$$ $$x = \pm y$$ Using our constraint, we have $x^2 +x^2 = 1$, so $x = \pm 1/ \sqrt{2}$. So, our possible maximum values are $(1/ \sqrt{2}, 1/ \sqrt{2}), (-1/ \sqrt{2}, 1/ \sqrt{2}), (1/ \sqrt{2}, -1/ \sqrt{2}), (-1/ \sqrt{2}, -1/ \sqrt{2})$. After checking values of $f$ at each of these points, we can conclude that the greatest $f$ value, $3/2$, occurs at $(1/ \sqrt{2}, 1/ \sqrt{2})$ and $(-1/ \sqrt{2}, -1/ \sqrt{2})$.

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Nicely explained! :) –  Mike Spivey Dec 4 '12 at 3:39