Determining Local Extrema with Lagrange

[cse63146]

[SOLVED] Determining Local Extrema with Lagrange

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

\nabla f(x,y,z) = \lambda g(x,y,z)

The Attempt at a Solution

So I did the partial derivatives for F and G:

\nabla f(x,y,z) = (8,4,-1) \nabla g(x,y,z) = (2x,2y,2x)

Now I use Lagrange to get

8 =2 \lambda x , 4 = 2\lambda y , -1 =2 \lambda z

Now I isolated \lambda = \frac{4}{x} = \frac{2}{y} =- \frac{1}{2z}

Now I re-write both y and z in terms of x and I get:

y = \frac{x}{2}, z =- \frac{x}{8}

Then I put them back into g(x,y,z) = g(x,x/2,x/8)

and get:

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}}

but the book says x = 8/3

Can anyone see what mistake did I make? Thank You.

 

[Office_Shredder]

Yeah I do. It's tricky. Let's rewind to... the step before

81x²/64 = 9

So dividing both sides by 81 we get

x²/64 = 1/9

Multiplying both sides by 64 we get

x² = 64/9

Try solving that one instead

 

[cse63146]

x = 8/3 which agrees with the book. Thank You.

but why couldn't I multiply 9 by 64, and then divide it by 81?

 

[gabbagabbahey]

x = 8/3 which agrees with the book. Thank You.

but why couldn't I multiply 9 by 64, and then divide it by 81?

You can, but 9*64/81=192/27=64/9 not 193/27