Question

Find two positive numbers \(x \space and\space  y\) such that \(x+y=60\) and \(xy^{3}\) is maximum.

Answer 1

The two numbers are \(x \space and\space  y\) such that\(x+y=60.\)

\(∴ y= 60-x\)

\(Let f(x) = xy^{3}.\)

\(f(x)= x(60-x)^{3}\)

\(f'(x)=(60-x)^{3}\)

\(f'(60-x)^{2}[60-x-3x]\)

\(=(60-x)^{2}(60-4x)\)

And,\(f''(x)=-2(60-x)-4(60-x)^{2}\)

\(=-2(60-x)[60-4x+2(60-x)]\)

\(=-2(60-x)(180-6x)\)

\(=-12(60-x)(30-x)\)

Now\(,f'(x)=0=x=60 \space or\space x=15\)

\(When \space x=60,f''(x)=0.\)

\(When\space x=15,f''(x)=-12(60-15)(30-15)=-12\times45\times15<0.\)

∴By second derivative test,\(x = 15\) is a point of local maxima of\( f.\)

Thus, function \(xy^{3}\) is maximum when \(x = 15\) and \(y = 60−15=45\). Hence, the required numbers are \(15 \space and \space45.\)