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Question Number 203933 by liuxinnan last updated on 02/Feb/24

$$a^2+b^2+2c^2=22$$$$provea+2b+c⩽11$$

Answered by York12 last updated on 02/Feb/24

$${(a+2b+c)}^2\stackrel{C-\mathrm{S}}{⩽}(a^2+b^2+2c^2)(1+4+\frac{1}{2})=121$$$$\iff a+2b+c⩽11.$$

Answered by witcher3 last updated on 02/Feb/24

$$\mathrm{a}+2\mathrm{b}+\mathrm{c}-\gamma ({\mathrm{a}}^2+{\mathrm{b}}^2+{2\mathrm{c}}^2-22)=\mathrm{f}(\mathrm{a},\mathrm{b},\mathrm{c},\gamma )$$$${\partial }_{\mathrm{a}}\mathrm{f}=0\Rightarrow 1-2\gamma \mathrm{a}=0$$$${\partial }_{\mathrm{b}}\mathrm{f}=0\Rightarrow 2-2\gamma \mathrm{b}=0$$$${\partial }_{\mathrm{c}}\mathrm{f}=0\Rightarrow 1-4\mathrm{c}\gamma =0$$$$\Rightarrow \mathrm{c}=\frac{1}{4\gamma },\mathrm{b}=\frac{1}{\gamma },\mathrm{a}=\frac{1}{2\gamma }$$$$\partial \gamma =0\Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2+{2\mathrm{c}}^2-22=0$$$$\Rightarrow \frac{1}{4}+1+\frac{1}{8}-22{\gamma }^2=0$$$$22{\gamma }^2=\frac{11}{8}\Rightarrow \gamma =\underset{-}{+}\frac{1}{4}$$$$\gamma =\frac{1}{4}\Rightarrow \mathrm{c}=1,\mathrm{b}=4,\mathrm{a}=2$$$$;\mathrm{f}(\mathrm{a},\mathrm{b},\mathrm{c},\gamma )⩽\mathrm{f}(2,4,1,\frac{1}{4})=2+8+1=11$$$$$$$$$$$$$$

Commented by York12 last updated on 02/Feb/24

$$\mathrm{is}\mathrm{not}\mathrm{that}\mathrm{lagrange}\mathrm{multipliers}$$

Commented by witcher3 last updated on 04/Feb/24

$$\mathrm{yes}$$

Commented by York12 last updated on 04/Feb/24

$$\mathrm{great}$$

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