From Rogawski ET section 14.8, exercise 5. Find the minimum and maximum values of the function subject to the given constraint f(x,y) = 15x2 + 3y2, 5x + 12y = 2 fmin = fmax = Solution f(x,y) = 15x2 + 3y2 df/dx = 30x + 6y dy/dx From the other equation dy/dx = -5 Hence 30x + 6y (-5) = 30x - 30y =0 Thus x = y This gives x = y = 2 / 17 fmin = 0.249.