Answer to Question #260596 in Calculus for vipul

Question

Show that the maximum value of a 2 b 2 c 2 on a sphere of radius r centered at the origin of a Cartesian abc-coordinate system in (r 2/3)3 .

Answer 1

Let $f(a,b,c)=a^2b^2c^2$

Where $a^2+b^2+c^2=r^2$

$\implies g(a,b,c)=a^2+b^2+c^2$

Use Lagrange multipliers to find the maximum value of $a^2b^2c^2$

$\nabla f(a,b,c)=\lambda\nabla g$

$\implies<2ab^2c^2,2a^2bc^2,2a^2b^2c>=\lambda<2a,2b,2c>$

$\implies ab^2c^2=a\lambda$ , $a^2bc^2=b\lambda$ , $a^2b^2c=c\lambda$

$\implies \lambda=b^2c^2$ , $\lambda=a^2c^2$ , $\lambda=a^2b^2$

$\implies b^2c^2=a^2c^2=a^2b^2\implies a=b=c$

Substitute a=b=c in $a^2+b^2+c^2=r^2$

$a=b=c=\frac{r}{\sqrt3}$

$(a^2b^2c^2)_{max}=(\frac{r^2}{3})^3$