Hi! Now i move on the the boundary points. Now i use the method of lagrange mult

Question

Hi!

Now i move on the the boundary points.

Now i use the method of lagrange multipliers.

This equation only gives that x=y=0, which is a point i already considered.

This gives me the point (2,2)

Now i examine all the points

So max of the function is 9 and the min is 0.

I dont know if these answers are correct. I also wondering if i am making the right assumption when the system of equations is inconsistence.

Thanks/Henrik

Find max&min; to the function f(x,y)=x^2-2xy+2y subject to the constraint D:{ (x,y):0 leq x leq 3,0 leq y leq 2} this is a rectangle. To find critical points if the interior of f i set triangle f=0 which gives me the point (1,1) Now i move on the the boundary points. Now i use the method of lagrange multipliers. since its a box i get 4 different constraint curves. g{1}(x,y)=x=3 ,g{2}(x,y)=x=0 , g{3}(x,y)=y=0 and g{4}(x,y)=y=2 for constraint g{1} i get this system of equations: which gives me the strange solution 2=6. Now my interpretation of the method of lagrange multipliers is that i am simply trying to find where the gradient of f is equal to the gradient of g times a scalar multiple. And from the above system of equation i can draw the conclusion that there is no such point along the constraint g{1}. the only points worth examining would be the points (3,0) and (3,2) along the constraint g{1}. Now onto constraint g{2} i get: this equation gives me that 2=0 which is impossible and hence i can rule out that there is any extreme points along the constraint g{2}. the only possible points along constraint curve g{2} would then be (0,0) and (0,2). now for constraint g{3}. This equation only gives that x=y=0, which is a point i already considered. Last constraint g{4}. This gives me the point (2,2) Now i examine all the points So max of the function is 9 and the min is 0.

Explanation / Answer

Yes, your solution is correct.

Your assumptions are correct since:

f(x,y)=x2-2xy+2y

Thus, first derivative fx=2(x-y), fy=2(1-x)

second derivative, fxx=2, fyy=0

and fxy=fyx=-2.

Now fx=0 gives x=y and fy=0 gives x=1.

Thus,  fx=0 and  fy=0 at (x,y)=(1,1)

And, if we calculate D= fxfy-( fxy)2

We get, D = -4 < 0

Thus, the critical point (x,y)=(1,1) is neither a maximum nor a minimum.

This follows that the function will have maximum and minimum along the perimeter of the rectangle

0<x<3, 0<y<2.

The subsequent steps done by you are correct.

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