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find the maximum value of 4x-3y-2z subject to 2x^2+3y^2+4z^2 =1

find the maximum value of 4x-3y-2z subject to 2x^2+3y^2+4z^2 =1

Grade:9

1 Answers

jagdish singh singh
173 Points
7 years ago
\hspace{-0.7 cm}$Using $\bf{C.S}$ Inequality.... \\\\$\bf{\left[(2\sqrt{2})^2+(-\sqrt{3})^2+(-1)^2\right]\cdot \left[(\sqrt{2}x)^2+(\sqrt{3}y)^2+(2z)^2\right]\geq (4x-3y-2z)^2}$\\\\\\So we get $\bf{(4x-3y-2z)^2\leq 12\times 1\Rightarrow (4x-3y-2z)\leq \sqrt{12}=2\sqrt{3}}.$\\\\\\and equality hold when $\bf{\frac{2\sqrt{2}}{\sqrt{2}x}=\frac{-\sqrt{3}}{\sqrt{3}y}=\frac{-1}{2z}.}$

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