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

taniska , 12 Years ago
Grade 12
anser 1 Answers
Rinkoo Gupta
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
Thanks & Regards
Rinkoo Gupta
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Last Activity: 12 Years ago
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