Ride The Saddle

Calculus Level pending

Consider the following class C 2 C^2 function f : R 2 → R f: \mathbb{R}^2 \to \mathbb{R} : f ( x , y ) = 4 x y 2 − 2 x 2 y − x + 1 f(x,y) = 4xy^2 - 2x^2y - x + 1

The x y xy -coordinates of the saddle points of f f are ( a , b ) (a, b) and ( a , − b ) (a, -b) .

Evaluate 1 a + b \dfrac{1}{a+b} .


The answer is 2.

Guilherme Dela Corte by Guilherme Dela Corte , 24, Brazil

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1 solution

Tom Engelsman
Feb 5, 2017

Setting the gradient of f ( x , y ) f(x,y) equal to zero gives:

∂ f ∂ x = 4 y 2 − 4 x y − 1 = 0 ; \frac{\partial f}{\partial x} = 4y^2 - 4xy - 1 = 0;

∂ f ∂ y = − 2 x 2 + 8 x y = 0. \frac{\partial f}{\partial y} = -2x^2 + 8xy = 0.

Upon observation in the second equation: − 2 x 2 + 8 x y = − 2 x ( x − 4 y ) = 0 ⇒ x = 0 , 4 y . -2x^2 + 8xy = -2x(x - 4y) = 0 \Rightarrow x = 0, 4y. Substituting these values into the first equation produces: 4 y 2 − 4 ( 0 ) ( y ) = 1 ⇒ y = ± 1 2 4y^2 - 4(0)(y) = 1 \Rightarrow y = \pm \frac{1}{2} and 4 y 2 − 4 ( 4 y ) ( y ) = 1 ⇒ − 12 y 2 = 1 4y^2 - 4(4y)(y) = 1 \Rightarrow -12y^2 = 1 , which are purely imaginary and are discounted. Thus the saddle points are ( a , b ) = ( 0 , ± 1 2 ) , (a,b) = (0, \pm \frac{1}{2}), and 1 a + b = 2 . \frac{1}{a + b} = \boxed{2}.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

How do you know that the point ( 0 , ± 1 2 ) (0, ±\frac{1}{2}) is a saddle point? Why can't it be a minimum or a maximum?

Stratos Giakoumakis - 4 years, 4 months ago

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After computing these points in the Hessian matrix, it appears just (0, -1/2) qualifies as a saddle point in that it has a positive first principal minor and a negative second principal minor. The point (0, 1/2) is negative for both of these same minors, which would be negative-definite => maximum. Good observation!

tom engelsman - 4 years, 4 months ago

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