Answer to Question #250512 in Calculus for Susan

Question #250512

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


1
Expert's answer
2022-02-01T15:41:03-0500

fx=8x,fy=18yf_x=8x,f_y=18y

gx=y,gy=xg_x=y,g_y=x


fx=λgx    8x=λyf_x=\lambda g_x \implies 8x=\lambda y

fy=λgy    18y=λxf_y=\lambda g_y \implies 18y=\lambda x


4x9y=yx    9y2=4x2\frac{4x}{9y}=\frac{y}{x}\implies 9y^2=4x^2


y=±2x/3y=\pm 2x/3


then:

2x2/3=6    x=±32x^2/3=6\implies x=\pm 3


solution:

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

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


f(x,y)=4(6/y)2+9y2f(x,y) = 4(6/y)^2 + 9y^2

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

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

so,

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


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