Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum
Solution and Explanation
Let one number be\( x\). Then, the other number is \((16 − x)\).
Let the sum of the cubes of these numbers be denoted by \(S(x)\). Then,
\(s(x)=x^{3}+(16-x)^{3}\)
\(s'(x)=3x^{2}-3(16-x)^{2},s"(x)=6x+6(16-x)\)
Now,\( s''(x)=0=3x^{2}-3(16-x)^{2}=0\)
\(x^{2}-(16-x)^{2}=0\)
\(x^{2}-256-x^{2}+32x=0\)
\(x=\frac{256}{32}=8\)
\(s''(8)=6(8)+6(16-8)=48+48=96>0\)
Now,
∴ By second derivative test, \(x = 8\) is the point of local minima of S.
Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8 and \(16 − 8 = 8\)
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