# Use Lagrange Multipliers to find the absolute extrema

Use Lagrange Multipliers to find the absolute extrema (if any) of:

$f(x,y) = 4x^2 + 9y^2$; subject to $2x +3y = 6$.

Using Lagrange I end up with one point: $(\frac{3}{2}, 1)$

I'm just not sure how to show if that point is a max or a min?

Note that $2x +3y = 6$ describes a line. Choose any other point on that line, say $(0, 2)$, and plug it into your original function. Is the value larger or smaller than when you plug your solution point into the function?

• Oops. Meant to say line; I'm rather sleep-deprived. Edited. – Kaj Hansen Apr 6 '14 at 21:20
• OK Awesome thanks. I see what you're saying. Don't know why I didn't think of that... – Dave Apr 7 '14 at 1:27

Another way to check the nature of the extremum in this situation is to substitute the constraint into the function. Thus, with $\ 3y \ = \ 6 \ - \ 2x \$ , we have

$$f(x,y) \ = \ 4x^2 \ + \ 9y^2$$

$$\rightarrow \ \ \phi(x) \ = \ 4x^2 \ + \ (6 \ - \ 2x)^2 \ = \ \ 4x^2 \ + \ 36 \ - \ 24x \ + \ 4x^2 \ = \ 4 \ ( \ 2x^2 \ - \ 6x \ + \ 9 \ )$$

$$= \ 4 \ ( \ 2 \ [ \ x^2 \ - \ 3x \ ] \ + \ 9 \ ) \ = \ 4 \ \left( \ 2 \ [ \ x^2 \ - \ 3x \ + \ \frac{9}{4} \ ] \ + \ 9 \ - \ \frac{9}{2} \ \right)$$

$$= \ 4 \ \left( \ 2 \ [ \ x \ - \ \frac{3}{2}\ ]^2 \ + \ \frac{9}{2} \ \right) \ = \ 8 \ [ \ x \ - \ \frac{3}{2}\ ]^2 \ + \ 18 \ \ .$$

We have thereby shown, since this function describes an "upward-opening" parabola, that the minimum of $\ \phi(x) \$ or $\ f(x,y) \$ is 18, which occurs at $\ x \ = \ \frac{3}{2} \ \Rightarrow \ y \ = \ 1 \ .$ A geometrical interpretation of the problem is that, by using the Lagrange-multiplier method, we are looking for level curves of the function $\ f(x,y) \$ which are just tangent to the constraint "curve", which is the line $\ 2x \ + \ 3y \ = 6 \$ . The level curves $\ 4x^2 \ + \ 9y^2 \ = \ C \$ are concentric ellipses, only one of which is large enough to "touch" the constraint line without passing through it. That ellipse is $\ 4x^2 \ + \ 9y^2 \ = \ 18 \$ ; ellipses with larger values of $\ C \$ cross the line at two points and can be made arbitrarily large. Thus, the Lagrange-multiplier method locates the absolute minimum of $\ f(x,y) \$ ; there is no absolute maximum.

Hint: another way to check: $$(4x^2+9y^2)(1+1) \geqslant (2x+3y)^2$$