Find-the-smallest-value-of-x-2-y-2-among-all-values-of-x-amp-y-satisfying-3x-y-20-

Question Number 141475 by bramlexs22 last updated on 19/May/21

F i n d t h e s m a l l e s t v a l u e o f

\sqrt{x^{2} + y^{2}} a m o n g a l l v a l u e s o f

x & y s a t i s f y i n g 3 x - y = 20

Answered by MJS_new last updated on 19/May/21

y = 3 x - 20

f (x) = \sqrt{10 (x^{2} - 12 x + 40)}

x = t + 6

f (t) = \sqrt{10 (t^{2} + 4)}

\min (f (t)) = 2 \sqrt{10}

Answered by mathmax by abdo last updated on 19/May/21

φ (x) = \sqrt{x^{2} + y^{2}} and y = 3 x - 20 \Rightarrow φ (x) = \sqrt{x^{2} + {(3 x - 20)}^{2}}

= \sqrt{x^{2} + {9 x}^{2} - 120 x + 400} = \sqrt{{10 x}^{2} - 120 x + 400}

φ^{^{'}} (x) = \frac{20 x - 120}{2 \sqrt{{10 x}^{2} - 120 x} + 400} = \frac{10 x - 60}{\sqrt{{10 x}^{2} - 120 x + 400}} = \frac{10 (x - 6)}{\sqrt{\dots .}}

φ^{^{'}} (x) = 0 \Rightarrow x = 6 and φ^{^{'}} (x) > 0 \Rightarrow x > 6

{φ^{^{'}} (x) < 0 \Rightarrow x < 6 \Rightarrow \inf φ (x) = φ (6) = \sqrt{6^{2} + (18 - 20})}^{2} = \sqrt{36 + 4}

= \sqrt{40} = 2 \sqrt{10}