Menu Close

Question-93166




Question Number 93166 by john santu last updated on 11/May/20
Answered by john santu last updated on 11/May/20
let put the right angle at B(0,0) , A(0,3)  C(4,0) . for a general right triangle  A(0,a), B(0,0),C(c,0)  we have X(x,y) ⇒T(x,y)=   AX^2 +BX^2 +CX^2   T= 3x^2 +3y^2 −2(cx+ay)+a^2 +c^2   we complete square   T= 3((x−(c/3))^2 −(c^2 /9)) + 3((y−(a/3))^2 −(a^2 /9))  + a^2 +c^2   that clearly has a minimum at   (x,y) = ((c/3),(a/3)) . For a = 3 , c = 4   we get T_(min)  = (2/3)(3^2 +4^2 ) = ((50)/3)  at X((4/3), 1)
$$\mathrm{let}\:\mathrm{put}\:\mathrm{the}\:\mathrm{right}\:\mathrm{angle}\:\mathrm{at}\:\mathrm{B}\left(\mathrm{0},\mathrm{0}\right)\:,\:\mathrm{A}\left(\mathrm{0},\mathrm{3}\right) \\ $$$$\mathrm{C}\left(\mathrm{4},\mathrm{0}\right)\:.\:\mathrm{for}\:\mathrm{a}\:\mathrm{general}\:\mathrm{right}\:\mathrm{triangle} \\ $$$$\mathrm{A}\left(\mathrm{0},\mathrm{a}\right),\:\mathrm{B}\left(\mathrm{0},\mathrm{0}\right),\mathrm{C}\left(\mathrm{c},\mathrm{0}\right) \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{X}\left({x},\mathrm{y}\right)\:\Rightarrow\mathrm{T}\left({x},\mathrm{y}\right)=\: \\ $$$$\mathrm{AX}^{\mathrm{2}} +\mathrm{BX}^{\mathrm{2}} +\mathrm{CX}^{\mathrm{2}} \\ $$$$\mathrm{T}=\:\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} −\mathrm{2}\left({cx}+{ay}\right)+{a}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$${we}\:{complete}\:{square}\: \\ $$$$\mathrm{T}=\:\mathrm{3}\left(\left({x}−\frac{{c}}{\mathrm{3}}\right)^{\mathrm{2}} −\frac{{c}^{\mathrm{2}} }{\mathrm{9}}\right)\:+\:\mathrm{3}\left(\left({y}−\frac{{a}}{\mathrm{3}}\right)^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{9}}\right) \\ $$$$+\:{a}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$${that}\:{clearly}\:{has}\:{a}\:{minimum}\:{at}\: \\ $$$$\left({x},\mathrm{y}\right)\:=\:\left(\frac{\mathrm{c}}{\mathrm{3}},\frac{{a}}{\mathrm{3}}\right)\:.\:\mathrm{For}\:{a}\:=\:\mathrm{3}\:,\:\mathrm{c}\:=\:\mathrm{4}\: \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{T}_{\mathrm{min}} \:=\:\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} \right)\:=\:\frac{\mathrm{50}}{\mathrm{3}} \\ $$$$\mathrm{at}\:\mathrm{X}\left(\frac{\mathrm{4}}{\mathrm{3}},\:\mathrm{1}\right)\: \\ $$
Commented by i jagooll last updated on 11/May/20
cool..==
$$\mathrm{cool}..== \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *