Question

Find two numbers whose sum is 12 such that the sum of the square of one plus 4 times the other is a maximum?

Answer 1

0 and 12

Explanation:

Let the first number be `x` (`0<= x<=12`). Then the second is `12-x`.

We need to maximize

`f(x) = x^2+4(12-x)^2`
`qquad quad quad = x^2+4(144-24x+x^2)`
`qquad qquad quad = 5x^2-96x+576`

Differentiating, we get

`(df)/dx = 10x-96`
`(d^2f)/dx^2 = 10`

The only local extremum is thus at `x = 96/10 = 9.6`. Since `(d^2f)/dx^2 >0` this is a minimum. The maximum must then be at one of the endpoints of the interval `[0,12]`. Since `f(0)=576>144=f(12)`, the maximum occurs when the first number is 0, and the second is 12.