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Question Number 180728 by CrispyXYZ last updated on 16/Nov/22

$$\mathrm{Given}x,y,z\in {\mathbb{R}}^{+}\mathrm{such}\mathrm{that}{x}^2-3xy+4y^2-z=0.$$$$\mathrm{when}\frac{xy}{z}\mathrm{reaches}\mathrm{its}\max \mathrm{value},\mathrm{find}\mathrm{the}\max \mathrm{value}\mathrm{of}$$$$\frac{2}{x}+\frac{1}{y}-\frac{2}{z}.$$

Answered by mr W last updated on 20/Nov/22

$$\frac{xy}{z}=maxmeans\frac{z}{xy}=min$$$$\frac{z}{xy}=\frac{x^2-3xy+4y^2}{xy}=\frac{x}{y}+\frac{4y}{x}-3⩾2\sqrt{\frac{x}{y}\times \frac{4y}{x}}-3=1$$$$min=1at\frac{x}{y}=\frac{4y}{x},i.e.\frac{x}{y}=2$$$$\frac{2}{x}+\frac{1}{y}-\frac{2}{z}=\frac{2}{x}+\frac{1}{y}-\frac{2}{x^2-3xy+4y^2}$$$$=\frac{2}{2y}+\frac{1}{y}-\frac{2}{4y^2-6y^2+4y^2}=\frac{2}{y}-\frac{1}{y^2}$$$$=-(\frac{1}{y^2}-\frac{2}{y}+1)+1=1-{(\frac{1}{y}-1)}^2⩽1$$$$i.e.{(\frac{2}{x}+\frac{1}{y}-\frac{2}{z})}_{max}=1$$

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