Question

Find two positive numbers \(x \space and\space  y\) such that their sum is \(35\) and the product \(x^{2} y^{5}\) is a maximum

Answer 1

Let one number be \(x\). Then, the other number is \(y = (35 − x)\).

Let\( p(x)=x^{2}y^{5}\).Then, we have:

\(p(x)=x^{2}(35-x)^{5}\)

\(p'(x)=2x(35-x)^{5}-5x^{2}(35-x)^{4}\)

\(=x(35-x)^{4}[2(35-x)-5x]\)

\(=x(35-x)^{4}(70-7x)\)

\(=7x(35-x)^{4}(10-x)\)

And\(.p''(x)=7(35-x)^{4}(10-x)+7x[-35-x)^{4}-4(35-x)^{3}(10-x)]\)

\(=7(35-x)^{4}(10-x)-7x(35-x)^{4}-28x(35-x)^{3}(10-x)\)

\(=7(35-x)^{3}[(35-x)(10-x)-x(35-x)-4x(10-x)]\)

\(=7(35-x)^{3}[350-45x+x^{2}-35x+x^{2}-40x+4x^{2}]\)

\(=7(35-x)^{3}(6x^{2}-120x+350)\)

Now,\(p(x)=0=x=0,x=35,x=10\)

When \(x = 35,\)\( f'(x)=f(x)=0\) \(and\) \(y = 35 − 35 = 0\). This will make the product\( x^{2} y^{5}\) equal to \(0\).

When \(x = 0, y = 35 − 0 = 35 \)and the product \( x^{2} y^{5}\) will be \(0\).

∴ \(x = 0 \space and \space x = 35\) cannot be the possible values of x. When \(x = 10\), we have:
\(P"(x)=7(35-10)^{3}(6\times 100-120\times10+350)\)
\(=7(25)^{3}(-250)<0\)

∴ By second derivative test, \(P(x)\) will be the maximum when \(x = 10\) and \(y = 35 − 10 = 25\). Hence, the required numbers are \(10 \space and \space 25\).