Find the critical point of the function f(x,y)=−(36x+6y^2+ln(|x+y|)) .

Question

Find the critical point of the function f(x,y)=−(36x+6y^2+ln(|x+y|)) .

Use the Second Derivative Test to determine whether it is a local min, max, saddle point or test fails

Answer 1

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The domain of the function consists of all the pairs (x,y) except those where x+y = 0.

We divide the domain into two subdomains.

Subdomain 1: x+y > 0

f(x, y) = -(36x + 6y² + ln(x+y))

∂f/∂x = -(36 + 1/(x+y))

∂f/∂y = -(12y + 1/(x+y))

On this subdomain both derivatives are defined everywhere.

Therefore critical point occurs when both derivatives are 0.

Setting

-(36 + 1/(x+y)) = 0

yields

x+y = -1/36

Any such point (x,y) would not lie in this subdomain.

Therefore the function has no critical point on this subdomain.

Subdomain 2: x+y < 0

f(x, y) = -(36x + 6y² + ln(-(x+y)))

∂f/∂x = -(36 + 1/(x+y))

∂f/∂y = -(12y + 1/(x+y))

On this subdomain both derivatives are defined everywhere.

Therefore critical point occurs when both derivatives are 0.

Setting ∂f/∂x = 0

-(36 + 1/(x+y)) = 0

yields

x+y = -1/36 (1)

Setting ∂f/∂y = 0

-(12y + 1/(x+y)) = 0

Substituting from (1)

-(12y - 36) = 0

Yields

y = 3

From (1)

x = -3 - 1/36

So the function has a critical point at (-3 - 1/36, 3).

To determine the nature of the critical point compute the second derivatives

∂²f/∂x² = 1/(x+y)²

∂²f/∂x∂y = ∂²f/∂y∂x = 1/(x+y)²

∂²f/dy² = -12 + 1/(x+y)²

The determinant

D = ∂²f/∂x² × ∂²f/dy² - ∂²f/∂x∂y × ∂²f/∂y∂x

= 1/(x+y)²(-12 + 1/(x+y)²) - (1/(x+y)²)²

= 1/(x+y)²(-12 + 1/(x+y)² - 1/(x+y)²)

= -12/(x+y)² < 0

The determinant is negative, so the critical point is a saddle point.