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If $x,y,z>0$ and $x^2+7y^2+16z^2=1\;,$ Then $\min(xy+yz+zx)$

$\bf{My\; Try::}$Using Multiplier Method:

$$f(x,y,z,\lambda) = (xy+yz+zx)-\lambda(x^2+7y^2+16z^2-1)$$

So $$f'(x,y,z,\lambda)_{\bf{y,z,\lambda=const.}} = y+z-2\lambda x$$

and $$f'(x,y,z,\lambda)_{\bf{x,z,\lambda=const.}}=x+z -32\lambda$$

and $$f'(x,y,z,\lambda)_{\bf{x,y,\lambda=const.}}=x+y-14\lambda$$

and $$f'(x,y,z,\lambda)_{\bf{x,y,z=const.}}=0-1(x^2+7y^2+14z^2)\lambda$$

My question is can we solve it without multiplier method, plz explain me, Thanks

juantheron
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    You did not take into account in the Lagrangian function $f$ the inequality constraints $x \geq 0, y \geq 0$ and $z \geq 0$. Moreover, your derivatives are wrong. – Alex Silva Nov 24 '16 at 11:11

2 Answers2

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Well, $xy+yz+zx>0$ obviously, but can be made as close to zero as needed say when $x\to 1, y=z\to 0$, so it doesn't have a minimum, and the infimum is $0$.

Macavity
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Your question has been answered but may be you mean Maximum of that expression.

It has not a minimum but has a Maximum!

This is my answer to Max question:

Obviousely $$ 2(x-3y)^2+3(y-2z)^2+(6z-x)^2 \ge 0 $$ after expansion we have $$ 3x^2+21y^2+48z^2 - 12(xy+yz+zx) \ge 0 $$ hence $$ x^2+7y^2+16z^2 \ge 4(xy+yz+zx)$$ according to assumption $$ x^2+7y^2+16z^2=1 $$ as a result $$ xy + yz + zx \le \dfrac{1}{4} $$ and equality occures at $$ (x,y,z) = (\dfrac{6}{4\sqrt{5}},\dfrac{2}{4\sqrt{5}},\dfrac{1}{4\sqrt{5}})$$ and finally $$ Max( xy + yz + zx ) =\dfrac{1}{4}$$

Little Prince
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