Question

The greatest value of $\mathrm{x}+\mathrm{y}+\mathrm{z}$for possible values of $\mathrm{x},\mathrm{y},\mathrm{z}$ such that ${\mathrm{x}}^3+{\mathrm{y}}^3+{\mathrm{z}}^3=1$ is

• A. $2^{\frac{1}{3}}$
• B. $3^{\frac{1}{3}}$
• C. $3^{\frac{2}{3}}$
• D. $3^{\frac{4}{3}}$

Answer 1

The correct option is C

$3^{\frac{2}{3}}$

The explanation for the correct answer.

Find the greatest value.

Given: ${\mathrm{x}}^3+{\mathrm{y}}^3+{\mathrm{z}}^3=1=\mathrm{k}$ (constant)

If ${{\mathrm{a}}_1}^{\mathrm{m}}+{{\mathrm{a}}_2}^{\mathrm{m}}+\dots ..{{\mathrm{a}}^{\mathrm{n}}}_{\mathrm{n}}=\mathrm{k},$ the greatest value of ${\mathrm{a}}_1+{\mathrm{a}}_2+\dots ..+{\mathrm{a}}_{\mathrm{n}}={\mathrm{n}}^{1-\frac{1}{\mathrm{m}}}{\mathrm{k}}^{\frac{1}{\mathrm{m}}}$ if $\mathrm{m}\notin (0,1)$ and ${\mathrm{a}}_1={\mathrm{a}}_2=\dots ..\mathrm{An}$

Here $\mathrm{m}=3,\mathrm{n}=3$

The greatest value $\mathrm{x}+\mathrm{y}+\mathrm{z}$is $3^{1-\frac{1}{3}}·{\left(1\right)}^{\frac{1}{3}}=3^{\frac{2}{3}}$

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