Find the extreme values of f subject to both constraints. f(x,y,z)=x^2+y^2+z

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Answered question

2021-09-15

Find the extreme values of f subject to both constraints. f(x,y,z)=x2+y2+z2; xy=1, y2z2=1

Answer & Explanation

Nathaniel Kramer

Nathaniel Kramer

Skilled2021-09-16Added 78 answers

f(x,y,z)=x2+y2+z2 
g(x,y,z)=xy=1 
h(x,y,z)=y2z2=1 
downf=λdowng+μdownh
fx=2x gx=1 hx=0 
fy=2y gy=1 hy=2y 
fz=2z gy=0 hz=2z 
fx=λgx+μhx 
fy=λgy+μhy 
fz=λgz+μhz 
2x=λ 
2y=λ+2μy 
2z=2μz 
2z+2μz=0 z(1+μ)=0 
z=0 μ=1 
2y=λ+2μy 2x=λ =λ+2(1)y x=λ2 =λ2y 
y=λ4 
xy=1 (λ2)(λ4)=1 
λ=43 So x=23 y=13 
If we plug y=13 into the other constraint y2z2=1 and try to solve for z we would get an imaginary number and because of this must reject the values x=23 and

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