Among all pairs of numbers (x,y) such that 6x+2y=30, find the pair for which the sum of squares, x^2+y^2, is minimum.

Answer 1

(4.5, 1.5)

6(4.5)+2(1.5)= 27+3= 30

4.5^2 + 1.5^2 = 20.25+2.25 = 22.5 = minimum x^2+y^2

(4,3) gives 6(4)+2(3) = 30

but 4^2+3^2 = 16+9=25>22.5

(5,0) gives 6(5)+2(0) = 30

but 5^2 +0^2 = 25>22.5

Notice 4.5 is exactly the midpoint between 4 and 5

Answer 2

want point on line to be minimum distance to point (0,0)

minimum distance is from (0,0) to line is the perpendicular line

(0,0) to y=-3x+15

the perpendicular line is y=(1/3)x

these two line intersect at x=4.5 and y=1.5

distance²=4.5²+1.5²=22;.5 is minimum

check with point to line distance fromula

d=|mx₀+b-y₀|/√(1+m²),,,,m=-3, b=15, x₀=0, y₀=0

d=15/√10

d²=22.5

Answer 3

Divide by 2, 3x + y = 30, so y = 30-3x

Substitute into x^2 + y^2 giving x^2 + (30-3x)^2 = x^2 + 900 - 180x + 9x^2

= 10x^2 -180x + 900

Let y = 10x^2 - 180x + 900

as a>0 this faces up

to minimise this we want to find the vertex

x = -b/2a = 180/20 = 9

y (in original system) = 30-3x = 30-27 = 3

So to minimise x^2 + y^2 choose x=9, y=3