Question

Two parts of $64$ such that the sum of their cubes is minimum will be-

• A. $44,20$
• B. $16,48$
• C. $32,32$
• D. $50,14$

Answer 1

Answer: (C)

$32,32$

Let one part be $x$.
Hence another part be $(64-x)$
Thus
Their cubes will be
$x^3+(64-x{)}^3=y$
Thus
$y^{\prime }=3x^2-3(64-x{)}^2=0$
Or
$x^2=(64-x{)}^2$
Or
$x=64-x$ and $x=-64+x$
Hence
$x=32$.
Hence both the parts are
$32,32$.