Find two numbers whose sum is \(24\) and whose product is as large as possible.
Let one number be x. Then, the other number is \((24 − x)\). Let \(P(x)\) denote the product of the two numbers. Thus, we have:
\(p(x) =x(24-x)24x-x^{2}\)
\(p'(x)=24-2x\)
\(p''(x)=-2\)
Now,
\(p'(x)=0⇒x=12\)
Also,
\(p''(12)=-2<0\)
∴By second derivative test, \(x=12\) is the point of local maxima of P. Hence, the product of the numbers is the maximum when the numbers are 12 and \(24−12=12.\)