Question

Find two numbers whose sum is 24 such that the sum of the square of one plus six times the other is a minimum.

Answer 1

Let $x=$ the first number.Then the second number will be $24-x.$

Consider the function

$$f(x)=x^2+6(24-x)$$

Find the first derivative with respect to $x$

$$f'(x)=(x^2+6(24-x))'=2x-6$$

Find the critical number(s)

$$f'(x)=0=>2x-6=0=>x=3$$

If $x<3,f'(x)<0, f(x)$ decreases.

If $x>3,f'(x)>0, f(x)$ increases.

The function $f(x)$ has a local minimum at $x=3.$

Since the function $f(x)$ has the only extremum, then the function $f(x)$ has the absolute minimum for $x\in \R$ at $x=3.$

The first number is $3.$ The second number is $21.$