Question Number 226901 by mr W last updated on 18/Dec/25
$${if}\:{x}+{y}=\mathrm{2}\:{with}\:{x},\:{y}\:>\mathrm{0},\:{find}\:{the} \\ $$$${minimum}\:{of}\:{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} }. \\ $$
Commented by fantastic2 last updated on 18/Dec/25
$$\mathrm{3} \\ $$
Answered by gregori last updated on 18/Dec/25
$$\:{f}\left({x}\right)=\:{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{3}\left(\mathrm{2}−{x}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:={x}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{12}} \\ $$$$\:{f}\:'\left({x}\right)=\:\mathrm{1}+\frac{\mathrm{4}{x}−\mathrm{6}}{\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{12}}} \\ $$$$\:\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{12}}\:=\:\mathrm{6}−\mathrm{4}{x}\:;\:\mathrm{6}−\mathrm{4}{x}>\mathrm{0}\: \\ $$$$\:\:{x}=\mathrm{1}\:{or}\:{x}=\mathrm{2}\: \\ $$$$\:{x}=\mathrm{2}\:\left({rejected}\right) \\ $$$$\:{minimum}\:\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} }\:\right)\:=\:\mathrm{3} \\ $$
Answered by TonyCWX last updated on 18/Dec/25
![y=2−x ⇒x+(√(x^2 +3(2−x)^2 )) ⇒x+(√(4x^2 −12x+12)) Let u=x+(√(4x^2 −12x+12)) ⇒u−x = (√(4x^2 −12x+12)) ⇒u^2 −2ux+x^2 =4x^2 −12x+12 ⇒−3x^2 +(12−2u)x+(u^2 −12)=0 D≥0 (12−2u)^2 −4(−3)(u^2 −12)≥0 144−48u+4u^2 +12u^2 −144≥0 16u^2 −48u≥0 16u(u−3)≥0 u≤0 [omitted] or u≥3 Minimum of u = 3 Minimum of x+(√(x^2 +3y^2 )) = 3](https://www.tinkutara.com/question/Q226906.png)
$${y}=\mathrm{2}−{x} \\ $$$$\Rightarrow{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{3}\left(\mathrm{2}−{x}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{x}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{12}} \\ $$$$ \\ $$$${Let}\:{u}={x}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{12}} \\ $$$$\Rightarrow{u}−{x}\:=\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{12}} \\ $$$$\Rightarrow{u}^{\mathrm{2}} −\mathrm{2}{ux}+{x}^{\mathrm{2}} =\mathrm{4}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{12} \\ $$$$\Rightarrow−\mathrm{3}{x}^{\mathrm{2}} +\left(\mathrm{12}−\mathrm{2}{u}\right){x}+\left({u}^{\mathrm{2}} −\mathrm{12}\right)=\mathrm{0} \\ $$$$ \\ $$$${D}\geqslant\mathrm{0} \\ $$$$\left(\mathrm{12}−\mathrm{2}{u}\right)^{\mathrm{2}} −\mathrm{4}\left(−\mathrm{3}\right)\left({u}^{\mathrm{2}} −\mathrm{12}\right)\geqslant\mathrm{0} \\ $$$$\mathrm{144}−\mathrm{48}{u}+\mathrm{4}{u}^{\mathrm{2}} +\mathrm{12}{u}^{\mathrm{2}} −\mathrm{144}\geqslant\mathrm{0} \\ $$$$\mathrm{16}{u}^{\mathrm{2}} −\mathrm{48}{u}\geqslant\mathrm{0} \\ $$$$\mathrm{16}{u}\left({u}−\mathrm{3}\right)\geqslant\mathrm{0} \\ $$$${u}\leqslant\mathrm{0}\:\left[{omitted}\right]\:{or}\:{u}\geqslant\mathrm{3} \\ $$$$ \\ $$$${Minimum}\:{of}\:{u}\:=\:\mathrm{3} \\ $$$${Minimum}\:{of}\:{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} }\:=\:\mathrm{3} \\ $$