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


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

$If a, b, c ⩾ 0 and \frac{x - 1}{4} = \frac{y - 3}{2} = \frac{z + 2}{3},$ $\min (x^{2} + y^{2} - z^{2}) = ?$
	Answered by liberty last updated on 23/Aug/21
	${ ((x=4k+1)),((y=2k+3)),((z=3k−2)) :} ⇒f(k)=(4k+1)^2 +(2k+3)^2 −(3k−2)^2 f(k)=4k^2 +12k+9+(7k−1)(k+3) f(k)=4k^2 +12k+9+(7k^2 +20k−3) f(k)=11k^2 +32k+6 f(k)_(min) when k=−((16)/(11)) f(k)_(min) = 11(((256)/(121)))−((32×16)/(11))+6 =((256+66−512)/(11)) = −((190)/(11))$
	${\begin{cases} x = 4 k + 1 \\ y = 2 k + 3 \\ z = 3 k - 2 \end{cases} \Rightarrow f (k) = {(4 k + 1)}^{2} + {(2 k + 3)}^{2} - {(3 k - 2)}^{2}$ $f (k) = {4 k}^{2} + 12 k + 9 + (7 k - 1) (k + 3)$ $f (k) = {4 k}^{2} + 12 k + 9 + ({7 k}^{2} + 20 k - 3)$ $f (k) = {11 k}^{2} + 32 k + 6$ $f {(k)}_{\min} when k = - \frac{16}{11}$ $f {(k)}_{\min} = 11 (\frac{256}{121}) - \frac{32 \times 16}{11} + 6$ $= \frac{256 + 66 - 512}{11} = - \frac{190}{11}$
	Commented by gloriousman last updated on 23/Aug/21

	$x, y, z are non - negative real numbers$
	Commented by iloveisrael last updated on 23/Aug/21
	$if k≥0 ⇒f(k)_(min) = 49 when { ((x=5)),((y= 5)),((z = 1)) :} ; x,y,z∈Z$
	$i f k ⩾ 0 \Rightarrow f {(k)}_{m i n} = 49$ $w h e n {\begin{cases} x = 5 \\ y = 5 \\ z = 1 \end{cases}; x, y, z \in Z$
	Commented by mr W last updated on 23/Aug/21

	$x = 4 k + 1 ⩾ 0 \Rightarrow k ⩾ - \frac{1}{4}$ $y = 2 k + 3 ⩾ 0 \Rightarrow k ⩾ - \frac{3}{2}$ $z = 3 k - 2 ⩾ 0 \Rightarrow k ⩾ \frac{2}{3}$ $\Rightarrow k ⩾ \frac{2}{3}$ $f (k) = 11 k^{2} + 32 k + 6$ $f {(k)}_{m i n} = f (\frac{2}{3}) = 11 \times {(\frac{2}{3})}^{2} + 32 \times \frac{2}{3} + 6$ $= \frac{290}{9} = 32.222$
	Commented by gloriousman last updated on 23/Aug/21

	$Nice!$
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