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Question Number 151782 by gloriousman last updated on 23/Aug/21

If a,b,c≥0 and ((x−1)/4)=((y−3)/2)=((z+2)/3), min(x^2 +y^2 −z^2 )=?

$$ \\ $$$$\mathrm{If}\:\mathrm{a},\mathrm{b},\mathrm{c}\geqslant\mathrm{0}\:\mathrm{and}\:\frac{\mathrm{x}−\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{y}−\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{z}+\mathrm{2}}{\mathrm{3}}, \\ $$$$\mathrm{min}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{z}^{\mathrm{2}} \right)=? \\ $$$$ \\ $$

Answered by liberty last updated on 23/Aug/21

{ ((x=4k+1)),((y=2k+3)),((z=3k−2)) :} ⇒f(k)=(4k+1)^2 +(2k+3)^2 −(3k−2)^2 f(k)=4k^2 +12k+9+(7k−1)(k+3) f(k)=4k^2 +12k+9+(7k^2 +20k−3) f(k)=11k^2 +32k+6 f(k)_(min) when k=−((16)/(11)) f(k)_(min) = 11(((256)/(121)))−((32×16)/(11))+6 =((256+66−512)/(11)) = −((190)/(11))

$$\:\begin{cases}{\mathrm{x}=\mathrm{4k}+\mathrm{1}}\\{\mathrm{y}=\mathrm{2k}+\mathrm{3}}\\{\mathrm{z}=\mathrm{3k}−\mathrm{2}}\end{cases}\:\Rightarrow\mathrm{f}\left(\mathrm{k}\right)=\left(\mathrm{4k}+\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2k}+\mathrm{3}\right)^{\mathrm{2}} −\left(\mathrm{3k}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\:\mathrm{f}\left(\mathrm{k}\right)=\mathrm{4k}^{\mathrm{2}} +\mathrm{12k}+\mathrm{9}+\left(\mathrm{7k}−\mathrm{1}\right)\left(\mathrm{k}+\mathrm{3}\right) \\ $$$$\:\mathrm{f}\left(\mathrm{k}\right)=\mathrm{4k}^{\mathrm{2}} +\mathrm{12k}+\mathrm{9}+\left(\mathrm{7k}^{\mathrm{2}} +\mathrm{20k}−\mathrm{3}\right) \\ $$$$\:\mathrm{f}\left(\mathrm{k}\right)=\mathrm{11k}^{\mathrm{2}} +\mathrm{32k}+\mathrm{6} \\ $$$$\mathrm{f}\left(\mathrm{k}\right)_{\mathrm{min}} \:\mathrm{when}\:\mathrm{k}=−\frac{\mathrm{16}}{\mathrm{11}} \\ $$$$\:\mathrm{f}\left(\mathrm{k}\right)_{\mathrm{min}} \:=\:\mathrm{11}\left(\frac{\mathrm{256}}{\mathrm{121}}\right)−\frac{\mathrm{32}×\mathrm{16}}{\mathrm{11}}+\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{256}+\mathrm{66}−\mathrm{512}}{\mathrm{11}}\:=\:−\frac{\mathrm{190}}{\mathrm{11}} \\ $$

Commented by gloriousman last updated on 23/Aug/21

x,y,z are non-negative real numbers

$$\mathrm{x},\mathrm{y},\mathrm{z}\:\mathrm{are}\:\mathrm{non}-\mathrm{negative}\:\mathrm{real}\:\mathrm{numbers} \\ $$

Commented by iloveisrael last updated on 23/Aug/21

if k≥0 ⇒f(k)_(min) = 49 when { ((x=5)),((y= 5)),((z = 1)) :} ; x,y,z∈Z

$${if}\:{k}\geqslant\mathrm{0}\:\Rightarrow{f}\left({k}\right)_{{min}} \:=\:\mathrm{49} \\ $$$$\:{when}\:\begin{cases}{{x}=\mathrm{5}}\\{{y}=\:\mathrm{5}}\\{{z}\:=\:\mathrm{1}}\end{cases}\:;\:{x},{y},{z}\in\mathbb{Z} \\ $$

Commented by mr W last updated on 23/Aug/21

x=4k+1≥0 ⇒k≥−(1/4) y=2k+3≥0 ⇒k≥−(3/2) z=3k−2≥0 ⇒k≥(2/3) ⇒k≥(2/3) f(k)=11k^2 +32k+6 f(k)_(min) =f((2/3))=11×((2/3))^2 +32×(2/3)+6 =((290)/9)=32.222

$${x}=\mathrm{4}{k}+\mathrm{1}\geqslant\mathrm{0}\:\Rightarrow{k}\geqslant−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${y}=\mathrm{2}{k}+\mathrm{3}\geqslant\mathrm{0}\:\Rightarrow{k}\geqslant−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${z}=\mathrm{3}{k}−\mathrm{2}\geqslant\mathrm{0}\:\Rightarrow{k}\geqslant\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow{k}\geqslant\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${f}\left({k}\right)=\mathrm{11}{k}^{\mathrm{2}} +\mathrm{32}{k}+\mathrm{6} \\ $$$${f}\left({k}\right)_{{min}} ={f}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)=\mathrm{11}×\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{32}×\frac{\mathrm{2}}{\mathrm{3}}+\mathrm{6} \\ $$$$=\frac{\mathrm{290}}{\mathrm{9}}=\mathrm{32}.\mathrm{222} \\ $$

Commented by gloriousman last updated on 23/Aug/21

Nice!

$$\mathrm{Nice}! \\ $$

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