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Question Number 139577 by bemath last updated on 29/Apr/21

Solve for real number(((x−1)^2 )/2)+(((y−2)^2 )/4)+(((z−3)^2 )/6)+3=∣x−1∣+∣y−2∣+∣z−3∣

$$\:\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{number}\frac{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}+\frac{\left(\mathrm{y}−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}}+\frac{\left(\mathrm{z}−\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{6}}+\mathrm{3}=\mid\mathrm{x}−\mathrm{1}\mid+\mid\mathrm{y}−\mathrm{2}\mid+\mid\mathrm{z}−\mathrm{3}\mid\: \\ $$

Answered by mr W last updated on 29/Apr/21

a=∣x−1∣≥0, b=∣y−2∣≥0, c=∣z−3∣≥0 (a^2 /2)+(b^2 /4)+(c^2 /6)+3=a+b+c (1/2)(a^2 −2a+1)+(1/4)(b^2 −4b+4)+(1/6)(c^2 −6c+9)+3−(1/2)−1−(3/2)=0 (1/2)(a−1)^2 +(1/4)(b−2)^2 +(1/6)(c−3)^2 =0 ⇒a=∣x−1∣=1 ⇒x=2 ⇒b=∣y−2∣=2 ⇒y=4 ⇒c=∣z−3∣=3 ⇒z=6

$${a}=\mid{x}−\mathrm{1}\mid\geqslant\mathrm{0},\:{b}=\mid{y}−\mathrm{2}\mid\geqslant\mathrm{0},\:{c}=\mid{z}−\mathrm{3}\mid\geqslant\mathrm{0} \\ $$$$\frac{{a}^{\mathrm{2}} }{\mathrm{2}}+\frac{{b}^{\mathrm{2}} }{\mathrm{4}}+\frac{{c}^{\mathrm{2}} }{\mathrm{6}}+\mathrm{3}={a}+{b}+{c} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left({a}^{\mathrm{2}} −\mathrm{2}{a}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{4}}\left({b}^{\mathrm{2}} −\mathrm{4}{b}+\mathrm{4}\right)+\frac{\mathrm{1}}{\mathrm{6}}\left({c}^{\mathrm{2}} −\mathrm{6}{c}+\mathrm{9}\right)+\mathrm{3}−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}−\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left({a}−\mathrm{1}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}\left({b}−\mathrm{2}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{6}}\left({c}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{a}=\mid{x}−\mathrm{1}\mid=\mathrm{1}\:\Rightarrow{x}=\mathrm{2} \\ $$$$\Rightarrow{b}=\mid{y}−\mathrm{2}\mid=\mathrm{2}\:\Rightarrow{y}=\mathrm{4} \\ $$$$\Rightarrow{c}=\mid{z}−\mathrm{3}\mid=\mathrm{3}\:\Rightarrow{z}=\mathrm{6} \\ $$

Commented by Rasheed.Sindhi last updated on 29/Apr/21

V Nice as usual!

$$\mathcal{V}\:\mathcal{N}{ice}\:{as}\:{usual}! \\ $$

Commented by mr W last updated on 29/Apr/21

thanks sir!

$${thanks}\:{sir}! \\ $$

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