Question

find maximum valueof the function f(x,y)=xye^-2x+3y​

Answer 1

Answer:

First look for critical points of f in the interior (x−1)2+(y−1)2<4. Do this by comparing the partial derivatives of f to zero. Next search for critical values on the boundary (x−1)2+(y−1)2=4 using Lagrange multipliers. Compute f in these points and, by putting everything together, you will find the maximum and minimum of f in the region.

EDIT: Computing the derivatives:

fx(x,y)=x(1−y)e−(x+y)fy(x,y)=y(1−x)e−(x+y)

Comparing both to zero we get:

x(1−y)=0y(1−x)=0

Simultaneous solutions are (0,0) and (1,1), which are both within the interior (but as we will see these points don't matter anyway). Computing f at these points gives 0 and e−2≈0.135 correspondingly.

Now to the boundary. It is given by g(x,y)=4 where g(x,y)=(x−1)2+(y−1)2 and the partial derivatives:

gx(x,y)=2(x−1)gy(x,y)=2(y−1)

Using the Lagrange multipliers method, we have to solve

g(x,y)=4fx(x,y)−λgx(x,y)=0fy(x,y)−λgy(x,y)=0

to find the critical points on the boundary. Substituting:

(x−1)2+(y−1)2=4y(1−x)e−(x+y)−λ2(x−1)=0x(1−y)e−(x+y)−λ2(y−1)=0

Assuming x,y≠1 we can divide the last two equations by (x−1) and (y−1) correspondingly, so they become:

ye−(x+y)=−2λxe−(x+y)=−2λ

which implies x=y. Substituting into g(x,y)=4, we have:

2(x−1)2=4⟹x=1±2–√

Corresponding values of f are

f(1−2–√,1−2–√)=(3−22–√)e−2+22√≈0.393f(1+2–√,1+2–√)=(3+22–√)e−2−22√≈0.0466

For the case where x=1 or y=1 we plug them directly into g(x,y)=4 to get

x=1⟹(y−1)2=4⟹y=−1,3y=1⟹(x−1)2=4⟹x=−1,3

Corresponding values of f are

f(−1,1)=f(1,−1)=−e0=−1f(3,1)=f(1,3)=3e−4≈0.0549

We conclude that the minimum value of −1 is attained at (−1,1) and (1,−1), and the maximum value of (3−22–√)e−2+22√ is attained at (1−2–√,1−2–√).