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Question Number 10045 by konen last updated on 21/Jan/17

$$\mathrm{y}⟩0$$$$\frac{{3\mathrm{x}}^2-3\mathrm{xy}+{\mathrm{y}}^2}{{\mathrm{y}}^2}=7\Rightarrow \max \left(\mathrm{x}\right)=?$$

Answered by mrW1 last updated on 21/Jan/17

$$3x^2-3xy+y^2=7y^2$$$$3x^2-3xy-6y^2=0$$$$x^2-yx-2y^2=0$$$$(x-2y)(x+y)=0$$$$x=(2y,-y)$$$$\Rightarrow max\left(x\right)=2y$$

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