# Thread: Simple question regarding use of Lagrange

1. ## Simple question regarding use of Lagrange

Hi I just hit Lagrange in my calculus course and I like it alot because it makes life easier than testing out ever possible solution (like we did the chapter before)

Anyway I've been trying to use Lagrange on this problem:

Find min max values of:
f(x,y) = xy-y^2 on the disc x^2+y^2 =< 1

The "normal" (not used to it yet) approach would be going parametric equation
x=cos(t)
y=sin(t) -pi<=t=<pi

However, is there any way possible to solve this cind of question using simply lagrange multipliers and clever reducing? I've not been able to reduce this problem in a way that would make things easier. If someone could show me how to think or how to attack a reducing problem like this i'd be very greatful.

Also; Is Lagrange always the best viable option for solving problems like these? I really dont like using trigonometry in problems like these because I feel like i'd have an much easier time screwing up my calculations in the end. Also it would be nice to learn how to solve one problem in two ways.

2. Originally Posted by Hanga
Hi I just hit Lagrange in my calculus course and I like it alot because it makes life easier than testing out ever possible solution (like we did the chapter before)

Anyway I've been trying to use Lagrange on this problem:

Find min max values of:
f(x,y) = xy-y^2 on the disc x^2+y^2 =< 1

The "normal" (not used to it yet) approach would be going parametric equation
x=cos(t)
y=sin(t) -pi<=t=<pi

I don't see anything "normal" using parametric equations to solve a min-max problem, specially if asked to do so using Lagrange multipliers.

However, is there any way possible to solve this cind of question using simply lagrange multipliers and clever reducing? I've not been able to reduce this problem in a way that would make things easier. If someone could show me how to think or how to attack a reducing problem like this i'd be very greatful.

Use normal, usual, every-day Lagrange multipliers to solve the problem! Perhaps you still don't know that good what're Lagrange multipliers? They do NOT usually use parametric equations...

Tonio

Also; Is Lagrange always the best viable option for solving problems like these? I really dont like using trigonometry in problems like these because I feel like i'd have an much easier time screwing up my calculations in the end. Also it would be nice to learn how to solve one problem in two ways.
.

3. Originally Posted by Hanga
Hi I just hit Lagrange in my calculus course and I like it alot because it makes life easier than testing out ever possible solution (like we did the chapter before)

Anyway I've been trying to use Lagrange on this problem:

Find min max values of:
f(x,y) = xy-y^2 on the disc x^2+y^2 =< 1

The "normal" (not used to it yet) approach would be going parametric equation
x=cos(t)
y=sin(t) -pi<=t=<pi
This is on the boundary of the disk, not the disk itself.

However, is there any way possible to solve this cind of question using simply lagrange multipliers and clever reducing? I've not been able to reduce this problem in a way that would make things easier. If someone could show me how to think or how to attack a reducing problem like this i'd be very greatful.

Also; Is Lagrange always the best viable option for solving problems like these? I really dont like using trigonometry in problems like these because I feel like i'd have an much easier time screwing up my calculations in the end. Also it would be nice to learn how to solve one problem in two ways.
$\displaystyle \nabla xy- y^2= y\vec{i}+ (x- 2y)\vec{j}= \vec{0}$ so y= 0 and x= 0 is the only critical point inside the disc.

But then you need to check on the boundary and I guess that is what you are asking about.
If you think of $\displaystyle x^2+ y^2= 1$ as a "level curve" of $\displaystyle f(x,y)= x^2+ y^2$ then a normal is given by $\displaystyle \nabla x^2+ y^2= 2x\vec{i}+ 2y\vec{j}$ (of course- a normal to the unit circle lies along a radius).
Max or min of the function on that circle will lie where the two gradients point in the same direction: $\displaystyle y\vec{i}+ (x- 2y)\vec{j}= \lambda (2x\vec{i}+ 2y\vec{j})$. That gives you two equations, $\displaystyle y= 2\lambda x$ and $\displaystyle x- 2y= 2\lambda y$ which, together with $\displaystyle x^2+ y^2= 1$ gives you three equations to solve for x, y, and $\displaystyle \lambda$.

Since you don't need to know $\displaystyle \lambda$, I recommend dividing one of the first two equations by the other to immediately eliminate $\displaystyle \lambda$:
$\displaystyle \frc{y}{x- 2y}= \frac{2\lambda x}{2\lambda y}= \frac{x}{y}$
That gives you $\displaystyle y^2= x^2- 2xy$ which, together with $\displaystyle x^2+ y^2= 1$ you can solve for x and y.