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Find maximum and minimum of $f(x,y,z)=x^2+y^2+z^2$ subject to constraints $x+2y+z=8$ and $x-y=4$

My try:I solved it by Lagrange multiplier method but I found one set of value as $x=52/11,y=8/11,z=20/11$ Then how can i find both maximum and minimum of $f$.

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    you can derive a function only in $x$ – 2017-02-21
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    sorry sir...i did not get you. – 2017-02-21
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    Use the constraints to eliminate $y$ and $z$ and express $f$ as a function of $x$ alone. – 2017-02-21
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    that's exactly what i said above – 2017-02-21
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    I know. I just tried to express it slightly different because OP apparently did not yet understand it :) – 2017-02-21

2 Answers 2

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We have $y=x-4$ and $z=8-2y-x=16-3x$. This gives (show it !):

$f(x,y,z)=11x^2-104x+272$

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    But this gives only the minimum value, sir. – 2017-02-21
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    this is not true, since we have no $$f(x,y,z)$$ please correct it – 2017-02-21
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    @Dr.SonnhardGraubner What is "not true" here, Graubner? – 2017-02-21
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from the first condition we obtain $$z=8-x-2y$$ and from the second one $$y=x-4$$ plugging this in our function $f(x,y,z)$ we get $$h(x)=11\,{x}^{2}-104\,x+272$$

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    $y=4-x$ is wrong (and therefore also the expression for $h$). – 2017-02-21
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    it is just corrected Martin – 2017-02-21