Let x and y be nonnegative real numbers. If xy = \frac{2}{5}, then find the minimum value of 6x^2 + \frac{3}{5}y^2 - 2x + 8y.
\(6x^2 + \frac{3}{5}y^2 - 2x + 8y \)
xy = 2/5 → y = 2 / (5x)
6x^2 + (3/5)(2 / (5x)^2 - 2x + 8(2 / (5x))
6x^2 + (12/125)x^(-2) - 2x + (16/5)x^(-1)
The graph shows that the minimum ≈ -26.525 when x ≈ -.06
+1679 
