Suppose f(x, y) = x2 + y2 - 6x - 8y + 5 How many critical points does f have in
ID: 3188091 • Letter: S
Question
Suppose f(x, y) = x2 + y2 - 6x - 8y + 5 How many critical points does f have in R2? If there is a local minimum, what is the value of the discriminant D at that point? If there is none, type N. If there is a local maximum, what is the value of the discriminant D at that point? If there is none, type N. If there is a saddle point, what is the value of the discriminant D at that point? If there is none: type N. What is the maximum value of f on R2? If there is none, type N. What is the minimum value of f on R2? If there is none, type N.Explanation / Answer
f(x,y) = x2 + y2 - 6x - 8y + 5 Calculate all the partial derivatives we need.
fx = 2x - 6 fy = 2y - 8 fxx = 2 fyy = 2 fxy =fyx = 0
To find the critical points, set fx and fy equal to 0.
2x - 6 = 0 2y - 8 = 0
x = 3, y = 4.
There is one critical point and it's (3, 4). Use the second partial derivative test to determine it's nature.
Evaluate the discriminant: D(x,y) = fxxfyy - (fxy)2 at (3, 4)
M(3, 4) = 4 - 0 = 4
Evaluate fxx at (x, y) = (3, 4)
fxx(3,4) = 2 > 0. (3, 4) is thus a min.
number of critical point: one.
The value of the discriminant at that point is: 4
The point is a minimum and has a value of : f(3,4) = 9 + 16 - 18 - 32 + 5 = -20