Extreme values(maximum,minimum is special cases), multivariable calculus

Question

We know that $ x,y,z >0$ and $x+y+z=1$

(i)For which $x,y,z$ will be $xyz$ maximum?

(ii)For which $x,y,z$ will be $x^2+y^2+z^2$ minimum?

I still have no idea, how to start the solution of these problems, can I maybe use somehow the Lagrange multiplier method?

For $(i)$ use https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means — lab bhattacharjee, May 14 '17 at 17:48
You could use Lagrange multipliers. Alternatively, you could uses standard inequalities like AM/GM or Cauchy-Schwarz. — Angina Seng, May 14 '17 at 17:48

score 2 · Accepted Answer · answered May 14 '17 at 17:49

2

for (i) use that $$\frac{x+y+z}{3}\geq \sqrt[3]{xyz}$$ for (ii) use that $$\frac{x+y+z}{3}\le \sqrt{\frac{x^2+y^2+z^2}{3}}$$

answered May 14 '17 at 17:49

Could you maybe help me how to use it exactly in this case? – Herrpeter May 14 '17 at 17:55
1

pluggiing $$x+y+z=1$$ in the inequalities – Dr. Sonnhard Graubner May 14 '17 at 17:56
So the maximum will be at $(1/3,1/3,1/3)$? – Herrpeter May 14 '17 at 18:05

score 2 · Answer 2 · answered May 14 '17 at 17:49

2

For $(ii)$

$$3(x^2+y^2+z^2)-(x+y+z)^2=(x-y)^2+(y-z)^2+(z-x)^2\ge0$$

answered May 14 '17 at 17:49

lab bhattacharjee

Can you give me just a hint, how can I use it exactly for $(ii)$? – Herrpeter May 14 '17 at 18:07
1

@Herrpeter, So we have $$3(x^2+y^2+z^2)\ge(x+y+z)^2=1^2$$ – lab bhattacharjee May 14 '17 at 18:08
So I get that $x^2+y^2+z^2+2xy+2xz+2yz=1$? – Herrpeter May 14 '17 at 18:11
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@Herrpeter, So we get $$x^2+y^2+z^2\ge\dfrac13$$ – lab bhattacharjee May 14 '17 at 18:11

2 Answers2