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.