(1 point) The function f(x, y)=xy(1−2x−6y) has 4 critical points. List them and select the type of critical point. Points should be entered as ordered pairs and listed in increasing lexicographic order. By that we mean that (x, y) comes before (z, w) if x

The points are (0,0) Minimum

(1/6, 1/18) Saddle point

Step-by-step explanation:

To solve the question we have

f(x,y) = xy(1-2x-6y)

Therefore, f(x,y) = xy-2x²y-6xy²

Finding the partial derivative of the above gives

fₓ = y - 4xy-6y²

= x - 2x²-12xy

Therefore, when the above equations are set to 0 we get

y - 4xy-6y² = 0.......(1)

x - 2x²-12xy  = 0......(2)

Dividing equation (1) and (2) by y and x respectively gives

1 - 4x - 6y = 0

1 - 2x - 12y = 0  

Solving gives

x = 0.167 = 1/6

y = 0.056 = 1/18

When we place x in equation (1) we have

y - 4(1/6)y-6y² = 0

y = 0 or y = 1/18

Similarly, x = 0 or 1/6

Therefore, the four critical points are

(0,0) (1/6,1/18)

When x = 0 and y = 0

f(x,y) = 0 Hence this is a minimum

When x = 1/6 and y = 1/18

f (x,y) = 1/324 Which is a  saddle point.

5x+10 is the correct one

hope this