Question

Divide 64 into two parts such that the sum of the cubes of two parts is minimum

Answer 1

let two parts are x and 64-x

then S= x3+(64-x)3

dS/dx=3x2+3(64-x)2(-1)

for maxima or minima ds/sx must be zero

so put ds/sx=0

3[x2-(64-x)2]=0

=> x=32

d2S/dx2=6x+3(2)(64-x)

put x= 32 , then d2s/dx2= 6(32)+6(64-32)=384 >0

so S is minima at x=32

so the parts are 32, 64-32 i.e 32, 32

Ans. 32,32