Question

If $x>0,y>0,x+y=24$ and $x^6+y^6$ is minimum then the numbers are

• A. $14,18$
• B. $12,20$
• C. $12,12$
• D. $16,12$

Answer 1

The correct option is C $12,12$

let, $f\left(x\right)=x^6+y^6$

Given $x+y=24$

$\therefore y=24-x$

$\therefore f\left(x\right)=x^6+{(24-x)}^6$

For the function to be minimum, $f^{\prime }\left(x\right)=0$

$\therefore \frac{d}{dx}[x^6+{(24-x)}^6]=0$

$\therefore 6x^5+6{(24-x)}^5\frac{d}{dx}(24-x)=0$

$\therefore 6x^5+6{(24-x)}^5(-1)=0$

$\therefore 6x^5-6{(24-x)}^5=0$

$\therefore 6[x^5-{(24-x)}^5]=0$

$\therefore x^5-{(24-x)}^5=0$

$\therefore x^5={(24-x)}^5$

Taking fifth root on both sides, we get,

$x=(24-x)$

$\therefore 2x=24$

$\therefore x=12$

Now, $x+y=24$ (Given)

$\therefore 12+y=24$

$\therefore y=12$