Among the pairs of number (x,y) such that 2x-y=12, find the pair for which the sum of the squares is a minimum. I have to solve for y and write the equation in

The sum of the squares equals x^2 + y^2 = x^2 + (2x - 12)^2 because we can substitute y = 2x - 12.

We get a quadratic function f(x) = 5x^2 - 48x + 144, whose minimum is attained at x = 48/(2*5) = 4.8 and equals 28.8. The second number is y = 2x - 12 = - -2.4.