Answer to Question #255054 in Calculus for Farhan

Question

Find the greatest and smallest values that the function f(x,y) = 4xy takes on the ellipse x^2 + 2^y2 = 1 by using the Method of Lagrange Multipliers.

Answer 1

We are asked for the extreme values of subject to the constraint $g(x, y)=x^2+y^2=1.$

Using Lagrange multipliers, we solve the equations $\nabla f=\lambda \nabla g$ and $g(x, y)=1$, which can be written as

$$f_x=\lambda g_x, f_y=\lambda g_y, g(x, y)=1$$

or as

$$4y=2\lambda x$$

$$4x=2\lambda y$$

$$x^2+y^2=1$$

From two first equations

$$\dfrac{y}{x}=\dfrac{x}{y}\ or\ x=y=0$$

But $x^2+y^2=1.$ Then we have

$$x^2=y^2=\dfrac{1}{2}$$

Therefore has possible extreme values at the points $(-\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2}), (-\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}),$

$(\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2}),$ and $(\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}).$ Evaluating at these four points, we find that

$$f(-\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2})=2$$

$$f(-\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2})=-2$$

$$f(\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2})=-2$$

$$f(\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2})=2$$

Therefore the maximum value of on the circle is $f(-\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2})=f(\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2})=2$ and the

minimum value is $f(-\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2})=f(\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2})=-2.$