How can I solve this using the Lagrange method?

Question

This is what I keep doing but the answer seems to be wrong every time.

Answer 1

What you should use for $ \ g(x) \ $ in the method is just $ \ x^2 \ + \ y^2 \ - \ 13 \ . $ Your gradient equations are correct. You conclude from them that

$$ \lambda \ = \ \frac{3}{2x} \ = \ - \frac{2}{2y} \ \ \Rightarrow \ \ y \ = \ - \frac{2}{3}x \ \ . $$

Insert this into the constraint equation $ \ \ x^2 \ + \ y^2 \ = \ 13 $ to find values $ \ x \ $ and $ \ y \ . $ Note that since this is equivalent to finding the intersections of a straight line with a circle, there will be two ordered-pair solutions. Compare the values of $ \ f(x,y) \ $ at these two points to find the maximum and minimum values of that function.

[Incidentally, all the numbers in the problem come out nicely... I might also mention that in problems with a linear function, subject to a constraint equation representing a curve with four-fold symmetry, the maxima and minima are at points on a line through the center of the curve (in this case, the origin).]

Answer 2

You have a small mistake: From $3 = \frac{6 \, \lambda}{2\,\lambda}$ you infer $3 = 3 \, \lambda$ instead of $3 = 3$.