If-a-b-c-0-and-x-1-4-y-3-2-z-2-3-min-x-2-y-2-z-2-

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

{\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

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!