What is the minimum value of the expression \(2x^2+3y^2+8x-24y+62\) for real x and y?
f (x, y) = 2x^2 + 3y^2 + 8x - 24y +62
Take partial derivatives in x and y and set these to 0
fx = 4x + 8 = 0 → x = -2
fy = 6y - 24 = 0 → y = 4
As WolframAlpha shows....(-2,4) is the min for this function.....and the min value = 6