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y-0-3x-2-3xy-y-2-y-2-7-max-x-




Question Number 10045 by konen last updated on 21/Jan/17
y⟩0  ((3x^2 −3xy+y^2 )/y^2 )=7 ⇒max(x)=?
$$\mathrm{y}\rangle\mathrm{0} \\ $$$$\frac{\mathrm{3x}^{\mathrm{2}} −\mathrm{3xy}+\mathrm{y}^{\mathrm{2}} }{\mathrm{y}^{\mathrm{2}} }=\mathrm{7}\:\Rightarrow\mathrm{max}\left(\mathrm{x}\right)=? \\ $$
Answered by mrW1 last updated on 21/Jan/17
3x^2 −3xy+y^2 =7y^2   3x^2 −3xy−6y^2 =0  x^2 −yx−2y^2 =0  (x−2y)(x+y)=0  x=(2y,−y)  ⇒max(x)=2y
$$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}{xy}+{y}^{\mathrm{2}} =\mathrm{7}{y}^{\mathrm{2}} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}{xy}−\mathrm{6}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{yx}−\mathrm{2}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({x}−\mathrm{2}{y}\right)\left({x}+{y}\right)=\mathrm{0} \\ $$$${x}=\left(\mathrm{2}{y},−{y}\right) \\ $$$$\Rightarrow{max}\left({x}\right)=\mathrm{2}{y} \\ $$

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