Suppose f(x,y) = XY(1-10X-5Y).

f(x,y) has 4 critical points. List them in increasing lexographic order. By that we mean that (x, y) comes before (z, w) if x<z or if x=z and y<w. Also, describe the type of critical point by typing MA if it is a local maximum, MI if it is a local minimim, and S if it is a saddle point.

First point (, ) of type ?
Second point (, ) of type ?
Third point (, ) of type ?
Fourth point (, ) of type ?

To find the critical points of the function f(x, y) = XY(1 - 10X - 5Y), we need to find all the points where the partial derivatives with respect to x and y are equal to zero. Let's calculate the partial derivatives first:

∂f/∂x = Y(1 - 10X - 5Y) - 10XY
∂f/∂y = X(1 - 10X - 5Y) - 5XY

Now, set both partial derivatives equal to zero and solve for x and y independently:

∂f/∂x = 0:
Y(1 - 10X - 5Y) - 10XY = 0
Simplify the equation: Y - 10XY - 5Y^2 - 10XY = 0
Rearrange the terms: -10XY - 10XY + Y - 5Y^2 = 0
Combine like terms: -20XY + Y - 5Y^2 = 0
Factor out common terms: Y(1 - 20X - 5Y) = 0

We have two cases to consider:
1. Y = 0
2. 1 - 20X - 5Y = 0

∂f/∂y = 0:
X(1 - 10X - 5Y) - 5XY = 0
Simplify the equation: X - 10X^2 - 5XY - 5XY = 0
Rearrange the terms: -10X^2 - 10XY + X = 0
Combine like terms: -10XY - 10XY - 10X^2 + X = 0
Factor out common terms: X(1 - 20X - 5Y) = 0

Again, we have two cases to consider:
1. X = 0
2. 1 - 20X - 5Y = 0

Now let's examine each case separately:

Case 1: Y = 0 and X = 0
If Y = 0, then from case 2, we have 1 - 20X - 5Y = 0, which simplifies to 1 - 20X = 0.
Solve for X: 20X = 1, X = 1/20
So the first critical point is (1/20, 0).

Case 2: 1 - 20X - 5Y = 0
If X = 0, we get 1 - 20X - 5Y = 0, which simplifies to 1 - 5Y = 0.
Solve for Y: 5Y = 1, Y = 1/5
So the second critical point is (0, 1/5).

Now let's consider the case where neither X nor Y is zero.

Case 3: 1 - 20X - 5Y = 0
From case 3, we have 1 - 5Y = 0, which simplifies to Y = 1/5.
Substituting Y = 1/5 into 1 - 20X - 5Y = 0, we get 1 - 20X - 5(1/5) = 0.
Simplify the equation: 1 - 20X - 1 = 0
Combine like terms: -20X = 0, X = 0.
So the third critical point is (0, 1/5).

Case 4: 1 - 20X - 5Y = 0
From case 4, we have X = 1/20.
Substituting X = 1/20 into 1 - 20X - 5Y = 0, we get 1 - 20(1/20) - 5Y = 0.
Simplify the equation: 1 - 1 - 5Y = 0
Combine like terms: -5Y = 0, Y = 0.
So the fourth critical point is (1/20, 0).

Now that we have all the critical points, let's list them in increasing lexicographic order:

First point: (0, 0) of type S (saddle point)
Second point: (0, 1/5) of type S (saddle point)
Third point: (1/20, 0) of type S (saddle point)
Fourth point: (1/20, 0) of type S (saddle point)

Note: In this case, all the critical points are saddle points.