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Multivariable Calculus · Multivariable Differentiation

Critical Points — Practice Problems and Worked Solutions

Classify peaks, valleys, and saddles.

The formulas you need here

Second derivative test

D = f_{xx} f_{y y} - f_{x y}^{2}

What it says: D > 0 with fₓₓ > 0 is a minimum, D > 0 with fₓₓ < 0 a maximum, D < 0 a saddle.
When to reach for it: Classifying a critical point of a surface.
Watch out: The cross term is squared and subtracted. D = 0 is inconclusive and needs another approach.

f = x^{2} - y^{2} : D = - 4 < 0

A saddle — a maximum one way and a minimum the other.

17 worked problems

Every step is generated and checked by a computer algebra system.

1.
Classify the critical point at the origin.
$f (x, y) = x^{2} + y^{2}$
Answer: $Local minimum$
Step-by-step solution
1. $D = f_{xx} f_{y y} - f_{x y}^{2} = 4$
2. $Local minimum$
2.
Classify the critical point at the origin.
$f (x, y) = - x^{2} - y^{2}$
Answer: $Local maximum$
Step-by-step solution
1. $D = f_{xx} f_{y y} - f_{x y}^{2} = 4$
2. $Local maximum$
3.
Classify the critical point at the origin.
$f (x, y) = x^{2} - y^{2}$
Answer: $Saddle point$
Step-by-step solution
1. $D = f_{xx} f_{y y} - f_{x y}^{2} = - 4$
2. $Saddle point$
4.
Classify the critical point at the origin.
$f (x, y) = 2 x^{2} + 3 y^{2}$
Answer: $Local minimum$
Step-by-step solution
1. $D = f_{xx} f_{y y} - f_{x y}^{2} = 24$
2. $Local minimum$
5.
Classify the critical point at the origin.
$f (x, y) = - 3 x^{2} - y^{2}$
Answer: $Local maximum$
Step-by-step solution
1. $D = f_{xx} f_{y y} - f_{x y}^{2} = 12$
2. $Local maximum$
6.
Classify the critical point at the origin.
$f (x, y) = x y$
Answer: $Saddle point$
Step-by-step solution
1. $D = f_{xx} f_{y y} - f_{x y}^{2} = - 1$
2. $Saddle point$
7.
Classify the critical point at the origin.
$f (x, y) = 3 x^{2} + y^{2}$
Answer: $Local minimum$
Step-by-step solution
1. $D = f_{xx} f_{y y} - f_{x y}^{2} = 12$
2. $Local minimum$
8.
Classify the critical point at the origin.
$f (x, y) = - 2 x^{2} - 5 y^{2}$
Answer: $Local maximum$
Step-by-step solution
1. $D = f_{xx} f_{y y} - f_{x y}^{2} = 40$
2. $Local maximum$
9.
Classify the critical point at the origin.
$f (x, y) = 2 x y$
Answer: $Saddle point$
Step-by-step solution
1. $D = f_{xx} f_{y y} - f_{x y}^{2} = - 4$
2. $Saddle point$
10.
Classify the critical point at the origin.
$f (x, y) = x^{2} + 4 y^{2}$
Answer: $Local minimum$
Step-by-step solution
1. $D = f_{xx} f_{y y} - f_{x y}^{2} = 16$
2. $Local minimum$
11.
Classify the critical point at the origin.
$f (x, y) = - x^{2} - 3 y^{2}$
Answer: $Local maximum$
Step-by-step solution
1. $D = f_{xx} f_{y y} - f_{x y}^{2} = 12$
2. $Local maximum$
12.
Classify the critical point at the origin.
$f (x, y) = 3 x^{2} - 2 y^{2}$
Answer: $Saddle point$
Step-by-step solution
1. $D = f_{xx} f_{y y} - f_{x y}^{2} = - 24$
2. $Saddle point$
13.
Classify the critical point at the origin.
$f (x, y) = 5 x^{2} + 2 y^{2}$
Answer: $Local minimum$
Step-by-step solution
1. $D = f_{xx} f_{y y} - f_{x y}^{2} = 40$
2. $Local minimum$
14.
Classify the critical point at the origin.
$f (x, y) = - 4 x^{2} - 4 y^{2}$
Answer: $Local maximum$
Step-by-step solution
1. $D = f_{xx} f_{y y} - f_{x y}^{2} = 64$
2. $Local maximum$
15.
Classify the critical point at the origin.
$f (x, y) = - x^{2} + y^{2}$
Answer: $Saddle point$
Step-by-step solution
1. $D = f_{xx} f_{y y} - f_{x y}^{2} = - 4$
2. $Saddle point$
16.
Classify the critical point at the origin.
$f (x, y) = x^{2} + 6 y^{2}$
Answer: $Local minimum$
Step-by-step solution
1. $D = f_{xx} f_{y y} - f_{x y}^{2} = 24$
2. $Local minimum$
17.
Classify the critical point at the origin.
$f (x, y) = - 6 x^{2} - y^{2}$
Answer: $Local maximum$
Step-by-step solution
1. $D = f_{xx} f_{y y} - f_{x y}^{2} = 24$
2. $Local maximum$

The formulas you need here

Second derivative test

17 worked problems

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