Suppose f ( x, y )= x^ 2+ y^ 2–2 x –10 y +2 (A) How many critical points does f

Question

Suppose f(x,y)=x^2+y^2–2x–10y+2

(A) How many critical points does f have in?

(B) If there is a local minimum, what is the value of the discriminant D at that point? If there is none, type "N".

(C) If there is a local maximum, what is the value of the discriminant D at that point? If there is none, type "N".

(D) If there is a saddle point, what is the value of the discriminant D at that point? If there is none, type "N".

(E) What is the maximum value of f on Extra close brace? If there is none, type "N".

(F) What is the minimum value of f on Extra close brace? If there is none, type "N".

Explanation / Answer

Given: f(x,y) = x^2 + y^2 - 2x - 10y + 2

First, finding the critical points.
f'(x) = 2x - 2, and f'(y) = 2y - 10 .

Setting these equal to 0

2x-2 = 0 2y-10 = 0

=> x=1 y=5

It yields the critical point (x,y) = (1,5) .

So number of critical points = 1

B) using the Second Derivative Test:
f''(x) = 2, f"(y) = 2, f'(xy) = 0
==> D = f"(x) *f"(y) - f'(xy)^2 = 2 * 2 - 0 = 4.

Since D(1,5) = 4 > 0 and f"(x) at (1,5) > 0,

we have a local minimum at (1,5).
In fact, this is a global minimum (via completing the square):
f(x,y) = x^2 + y^2 - 2x - 10y + 2
= (x^2 - 2x) + (y^2 - 10y) + 2
= (x^2 - 2x + 1) + (y^2 - 10y + 25) + 2 -1 -25
= (x - 1)^2 + (y -5)^2 - 24

We have a global minimum of -24 where x - 1 = 0 and y - 5 = 0

<==> (x, y) = (1, 5)

C) N

D) N

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