Question 1136263
Among all pairs of numbers (x,y) such that 8x+2y=18, find the pair for which the sum of squares, x2+y2, is minimum.
 Write your answers as fractions reduced to lowest terms.
:
8x + 2y = 18
simplify, divide by 2
4x + y = 9
y = -4x + 9
:
sum = x^2 + y^2
replace y with (-4x+9)
sum = x^2 + (-4x+9)^2
FOIL (-4x+9)(-4x+9)
sum = x^2 + 16x^2 - 36x - 36x + 81
sum = 17x^2 - 72x + 81
min occurs on the axis of symmetry; x = -b/(2a)
x = {{{(-(-72))/(2*17)}}}
x = {{{72/34}}}
x = {{{36/17}}}
:
find y
y = {{{17(36/17)^2 - 72{{{36/17}}} + 81
y = 17({{{1296/289) - {{{2592/17}}} + 81
y = {{{1296/17}}} - {{{2592/17}}} + {{{1377/17}}}
y = {{{81/17}}}
:
Minimum of x^2 + y^2: x = 36/17; y = 81/17

Looks pretty wild, I hope this is right, let me know. ankor@att.net