Question

Find two numbers whose sum is 40, such that the sum of the square of one number plus 6 times the other number is a minimum.

Answer 1

let that number are a and b respectively.
according information above, we get the equation :
x+y = 40

asking the minimum value of K (assumed) = x^2 + 6y

from the first equation x+y = 40 ----> y = 40 - x. subtitute this into K, get :
K = x^2 + 6y
K = x^2 + 6(40-x)
K = x^2 - 6x + 240

K has the minimum value when x = -b/2a
a = coefficient of x^2 while b = coefficient of x.
so, x = -b/2a = -(-6)/(2*1) = 3

because y = 40 - x, subtitute x = 3 into y we get
y = 40 - 3 = 37

thus, they are x = 3 and y =37