Question

Among all pairs of numbers (x,y)such that 6x+2y=36 find the pair for which the sum of squares, x^2+y^2 is minimum.

Answer 1

First, we simplify 6x+2y=36 into 3x+y=18 by dividing by 2. This means that y=-3x+18.


The sum $x^2+y^2$ can be written as: $x^2+y^2=(x+y)^2-2xy$, 

from the binomial expansion formula: $(x+y)^2=x^2+2xy+y^2$.

Thus, substituting y=-3x+18 and simplifying we have

$x^2+y^2=(x+(-3x+18))^2-2x(-3x+18)$$=(-2x+18)^2+6x^2-36x=4x^2-72x+18^2+6x^2-36x$$=10x^2-108x+18^2$.

This is a parabola which opens upwards (the coefficient of x^2 is positive), so its minimum is at the vertex. To find x, we apply the formula -b/2a. Substituting b=-108, a=10, we find that x is 108/20=5.4.


At x=5.4, the expression $10x^2-108x+18^2$, which is equivalent to $x^2+y^2$, takes it smallest value.

Substituting, we would find $10x^2-108x+18^2=10(5.4)^2-108(5.4)+18^2=291.6-583.2+324$=32.4 This is the smallest value of the expression. 


For x=5.4, y=-3x+18=-3(5.4)+18=1.8.



Answer:   (5.4, 1.8)