To find: The alternate solution to the exercise with the help of Lagrange Multiplier. x+2y+3z=6

Tahmid Knox 2021-02-19 Answered
To find: The alternate solution to the exercise with the help of Lagrange Multiplier. x+2y+3z=6
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escumantsu
Answered 2021-02-20 Author has 98 answers

Method used: In order to find the maximum and minimum values of a function f(x, y, z) that is subjected to the constraintg(x,y,z)=k such that g0

a) Find the values x, y, z and λ when f(x,y,z)=λg(x,y,z) and g(x,y,z)=k

b) Compute the values at the points from step a and find the maxomum and minimum values of f.

Calculation: It is given that the plane is x+2y+3z=6 that lies in the first octant. Consider a point (x,y,z) on the plane. Then, the function becomes f(x,y,z)=xyz. Here, g(x,y)=x+2y+3z.

Fint fx as fx(x,y,z)=yz,fy as fy(x,y,z)=xz,fz as fz(x,y,z)=xy,gx as gx(x,y,z)=1,gy as gy(x,y,z)=2andgzasgz(x,y,z)=3 Thus, the equations are obtained as follows. fx(x,y,z)=λgx(x,y,z)
yz=λ(1)
yz=λ Solve the equation fy(x,y,z)=xz as shown below. fy(x,y,z)=λgy(x,y,z)\
xz=λ(2)
xz=2λ
xz2=λ Solve the equation fz(x,y,z)=xy as follows. fz(x,y,z)=λgz(x,y,z)
xy=λ(3)
xy=3λ
xy3=λ Then, yz=xz2=xy3=λ Consider two terms at a time, take yz=xz2 that gives yz=xz2
y=x2

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