Answer to Question #250512 in Calculus for Susan

Question

Use Lagrange multipliers to find the maximum and minimum values of the function ( if they exist) subject to the specified constraint.

f(x,y) = 4x^2 + 9y^2 on the hyperbola xy = 6

Answer 1

$f_x=8x,f_y=18y$

$g_x=y,g_y=x$

$f_x=\lambda g_x \implies 8x=\lambda y$

$f_y=\lambda g_y \implies 18y=\lambda x$

$\frac{4x}{9y}=\frac{y}{x}\implies 9y^2=4x^2$

$y=\pm 2x/3$

then:

$2x^2/3=6\implies x=\pm 3$

solution:

$(3,2),(3,-2),(-3,2),(-3,-2)$

$f(3,2)=f(3,-2)=f(-3,2)=f(-3,-2)=72$

$f(x,y) = 4(6/y)^2 + 9y^2$

if we take, for example, y =1we get:

$f(x,1)=153>72$

so,

f(3,2)=f(3,-2)=f(-3,2)=f(-3,-2)=72 are minimums