Also use the method of Lagrange Multipliers to find the maximum and minimum valu

Question

Also use the method of Lagrange Multipliers to find the maximum and minimum values of the following functions f subject to the constraint g=c.
1. f(x,y)=x^2+y^2, g(x,y)=xy=1
2. f(x,y,z)=2x+6y+10z, g(x,y,z)=x^2+y^2+z^2=35
3. f(x,y,z)=xyz, g(x,y,z)=x^2+2y^2+3z^2=6
please show steps thanks

Explanation / Answer

1) partial derivatives df/dx=2x dg/dx=y df/dy=2y dg/dy=x thus we get 2x=yc and 2y=xc thus putting this in function g(x,y) we get x=y thus x^2=1 we get (1,1) and (-1,-1) the two point shere... we check d^2f/dx^2= 2 and d^2f/dy^2 =2 d^ 2f/dxdy =0 thus D=2*2-0 =4 > 0 thus (1,1) and (-1,-1) are the local minima here with max value =2 2) df/dx=2 dg/dx=2x df/dy=6 dg/dy=2y df/dy=10 dg/dz=2z from a) we have x=c b)y=3c c) z=5c thus y=3x and z=5x thus fuction g(x,y,z) becomes 35x^2= 35 thus x= 1 or -1 thus y= 3 or -3 and z= 5 or -5 we get two points (1,3,5) and(-1,-3,-5) these two are global maxima thus maxi mum value is 70 and minimum is -70

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