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[SOLVED] Determining Local Extrema with Lagrange
Find local extram of f(x,y,z) = 8x+4y-z with constraint g(x,y,z) = x^2 + y^2 + z^2 = 9
[tex]\nabla f(x,y,z) = \lambda g(x,y,z)[/tex]
So I did the partial derivatives for F and G:
[tex]\nabla f(x,y,z) = (8,4,-1)[/tex] [tex]\nabla g(x,y,z) = (2x,2y,2x)[/tex]
Now I use Lagrange to get
[tex]8 =2 \lambda x , 4 = 2\lambda y , -1 =2 \lambda z[/tex]
Now I isolated [tex]\lambda = \frac{4}{x} = \frac{2}{y} =- \frac{1}{2z}[/tex]
Now I re-write both y and z in terms of x and I get:
[tex]y = \frac{x}{2}, z =- \frac{x}{8}[/tex]
Then I put them back into g(x,y,z) = g(x,x/2,x/8)
and get:
[tex]g(x,x/2,x/8) = x^2 + \frac{x^2}{4} +\frac{x^2}{64} = 9 \Rightarrow \frac{64x^2 + 16x^2 + x^2}{64} = 9 \Rightarrow \frac{81x^2}{64} = 9 \Rightarrow x = \sqrt{\frac{193}{27}}[/tex]
but the book says x = 8/3
Can anyone see what mistake did I make? Thank You.
Homework Statement
Find local extram of f(x,y,z) = 8x+4y-z with constraint g(x,y,z) = x^2 + y^2 + z^2 = 9
Homework Equations
[tex]\nabla f(x,y,z) = \lambda g(x,y,z)[/tex]
The Attempt at a Solution
So I did the partial derivatives for F and G:
[tex]\nabla f(x,y,z) = (8,4,-1)[/tex] [tex]\nabla g(x,y,z) = (2x,2y,2x)[/tex]
Now I use Lagrange to get
[tex]8 =2 \lambda x , 4 = 2\lambda y , -1 =2 \lambda z[/tex]
Now I isolated [tex]\lambda = \frac{4}{x} = \frac{2}{y} =- \frac{1}{2z}[/tex]
Now I re-write both y and z in terms of x and I get:
[tex]y = \frac{x}{2}, z =- \frac{x}{8}[/tex]
Then I put them back into g(x,y,z) = g(x,x/2,x/8)
and get:
[tex]g(x,x/2,x/8) = x^2 + \frac{x^2}{4} +\frac{x^2}{64} = 9 \Rightarrow \frac{64x^2 + 16x^2 + x^2}{64} = 9 \Rightarrow \frac{81x^2}{64} = 9 \Rightarrow x = \sqrt{\frac{193}{27}}[/tex]
but the book says x = 8/3
Can anyone see what mistake did I make? Thank You.
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