Question

Find two positive numbers x and y such that their sum is 16 and sum of whose cubes is minimum.

Answer 1

Toolbox:

ddx(x.y)=x.ddx(y)+y.ddx(x)

Given

Sum of the No. = 16

Let sum of cubes be S

Let two numbers be x and 16−x

Sum of cubes S=x3+(16−x)3

Differentiating with respect to x we get

dSdx=3x2+3(16−x)2.(−1)

=3x2−3(16−x)2.(−1)

=3(32x−256)

Step 2:

For minimum value

dSdx=0

3(32x−256)=0

x=25632=8

Also d2Sdx2=96>0

⇒S is minimum at x=8

Hence the required numbers are 8 & (16−8)⇒ 8 & 8