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

1 Answer
Jun 19, 2018

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.