Answer to Question #332883 in Calculus for Munif

Question

Use Lagrange Multipliers to find the maximum and minimum values of f(x,y)=xy

 subject to the constraint 4x^2+8y^2=16

.

Answer 1

$f(x,y)=xy, \ \ 4x^2+8y^2=16.$

Let $g(x,y)=4x^2+8y^2.$

We have $f_x=y,\ \ f_y=x$ and $g_x=8x,\ \ g_y=16y.$

$\begin{cases} f_x=\lambda g_x \\ f_y=\lambda g_y \\ 4x^2+8y^2=16 \end{cases} \quad \begin{cases} y=8\lambda x \\ x=16\lambda y \\ 4x^2+8y^2=16 \end{cases}$

$x=16\lambda y=16\lambda \cdot8\lambda x=128\lambda ^2x$

$x\neq 0\ \Rightarrow \ 128\lambda^2=1,\ \ \lambda =\pm \tfrac{1}{8\sqrt{2}}$

$4x^2+8y^2=4x^2+8\cdot (8\lambda x)^2=8x^2=16$

$x=\pm\sqrt{2},$ $y=\pm1$

Maximum values: $f(\sqrt{2},1)=f(-\sqrt{2},-1)=\sqrt{2}$

Minimum values:

$f(-\sqrt{2},1)=f(\sqrt{2},-1)=-\sqrt{2}$

Answer: maximum value is $\sqrt{2}$ , minimum value is $-\sqrt{2}$ .