Question

The least value of $\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}+\mathrm{f}+\mathrm{g}+\mathrm{h}$ for possible values of $\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d},\mathrm{e},\mathrm{f},\mathrm{g},$ and $\mathrm{h}$ satisfying the condition ${\mathrm{a}}^{\frac{2}{3}}+{\mathrm{b}}^{\frac{2}{3}}+{\mathrm{c}}^{\frac{2}{3}}+{\mathrm{d}}^{\frac{2}{3}}+{\mathrm{e}}^{\frac{2}{3}}+{\mathrm{f}}^{\frac{2}{3}}+{\mathrm{g}}^{\frac{2}{3}}+{\mathrm{h}}^{\frac{2}{3}}=4$ is

• A. $\sqrt{2}$
• B. $2\sqrt{2}$
• C. $3\sqrt{2}$
• D. $4\sqrt{2}$

Answer 1

The correct option is B

$2\sqrt{2}$

Explanation of the correct option.

Compute the required value.

Given : ${\mathrm{a}}^{\frac{2}{3}}+{\mathrm{b}}^{\frac{2}{3}}+{\mathrm{c}}^{\frac{2}{3}}+{\mathrm{d}}^{\frac{2}{3}}+{\mathrm{e}}^{\frac{2}{3}}+{\mathrm{f}}^{\frac{2}{3}}+{\mathrm{g}}^{\frac{2}{3}}+{\mathrm{h}}^{\frac{2}{3}}=4=\mathrm{Constant}$

Here $\mathrm{m}=\frac{2}{3}$ and $\mathrm{n}=8,\left[0<\mathrm{m}<1\right]$

Therefore the least value of $\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}+\mathrm{f}+\mathrm{g}+\mathrm{h}$ $=8^{1-\frac{3}{2}}\times 4^{\frac{3}{2}}$

The least value of $\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}+\mathrm{f}+\mathrm{g}+\mathrm{h}$ $={\left(2^3\right)}^{-\frac{1}{2}}\times {\left(2^2\right)}^{\frac{3}{2}}$

The least value of $\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}+\mathrm{f}+\mathrm{g}+\mathrm{h}$ $=2\sqrt{2}$

Hence option $\left(B\right)$ is the correct option.