SME
Calculus III (Multivariable) · Problem type
Classify a critical point
Use this for “Find the critical point of f(x, y) = x² + xy + y² − 3x and say whether it is a maximum, a minimum or a saddle point.”
Diagram
f(x, y) = x² + xy + y² − 3x.
∇f = 0 at (2, −1), where D = 2 × 2 − 1² = 3 > 0 and ∂²f/∂x² = 2: a minimum, f = −3.
Heights are squeezed to fit: one unit of z is drawn 0.33 as long as one of x or y.
This is an example. Type your own values to start a new problem.
f(x,
y)
=
a
x²
+
b
xy
+
c
y²
+
(
d
)
x
+
e
y
f(
x
,
y
)
=
f(x, y)
D
Discriminant 4ac − b²
worked out (not typed)
K
Kind of point
worked out (not typed)
Formulas
D
=
4
×
a
×
c
−
b
2
3
=
4
×
1
×
1
−
1
2
x
=
b
×
e
− 2 ×
c
×
d
D
2
=
1 × 0 − 2 × 1 × (−3)
3
y
=
b
×
d
− 2 ×
a
×
e
D
−1
=
1 × (−3) − 2 × 1 × 0
3
f
(
x
,
y
)
=
a
×
x
2
+
b
×
x
×
y
+
c
×
y
2
+
d
×
x
+
e
×
y
−3
=
1
×
2
2
+
1
×
2
×
(−1)
+
1
×
(−1)
2
+
(−3)
×
2
+
0
×
(−1)
K
=
minimum
if
D
>
0
and
a
>
0,
maximum
if
D
>
0
and
a
<
0,
saddle
point
if
D
<
0
3 > 0 and 1 > 0, so minimum
Assumptions
•
f(x, y) = ax² + bxy + cy² + dx + ey. At a critical point ∂f/∂x = 0 and ∂f/∂y = 0.
•
Second derivative test: D = fₓₓ·fᵧᵧ − (fₓᵧ)², here (2a)(2c) − b² = 4ac − b².
•
D > 0 with a > 0 is a minimum, D > 0 with a < 0 a maximum, and D < 0 a saddle point.
•
With D = 0 there is no single critical point, and the test says nothing.
Step-by-step
WE KNOW
a = 1, b = 1, c = 1, d = −3, e = 0
FIND
discriminant 4ac − b² (D), critical point, x (x), critical point, y (y), height there (f(x, y)), kind of point (K)
Step 1 · Find discriminant 4ac − b² (D)
Use D = 4 × a × c − b²
The
second
partials
are
fₓₓ
=
2
a
,
fᵧᵧ
=
2
c
and
fₓᵧ
=
b
,
so
D
=
(
2
a
)(
2
c
)
−
b
2
.
D
=
4
×
1
×
1
−
1
2
D
=
4
−
1
D
=
3
Step 2 · Find critical point, x (x)
Use x = (b × e − 2 × c × d) ÷ D
Set
both
partials
to
0:
2ax
+
by
=
−
d
and
bx
+
2cy
=
−
e
.
Solve
the
pair
by
Cramer’s
rule.
x
=
1 × 0 − 2 × 1 × (−3)
3
2
x
+
y
=
3
x
+
2
y
=
0
x
=
2
Step 3 · Find critical point, y (y)
Use y = (b × d − 2 × a × e) ÷ D
The
same
pair
of
equations
gives
y
.
y
=
1 × (−3) − 2 × 1 × 0
3
y
=
−
3
3
y
=
−1
Step 4 · Find height there (f(x, y))
Use f(x, y) = a × x² + b × x × y + c × y² + d × x + e × y
Put the critical point into f for its height.
f
(
x
,
y
)
=
1
×
2
2
+
1
×
2
×
(−1)
+
1
×
(−1)
2
+
(−3)
×
2
+
0
×
(−1)
f
(
x
,
y
)
=
4
−
2
+
1
−
6
f
(
x
,
y
)
=
−3
Step 5 · Find kind of point (K)
Use K = minimum if D > 0 and a > 0, maximum if D > 0 and a < 0, saddle point if D < 0
With
D
>
0
the
surface
curves
the
same
way
in
every
direction,
and
fₓₓ
=
2
a
>
0
makes
it
curve
up:
a
minimum.
3 > 0 and 1 > 0: minimum
K
=
minimum
if
3
>
0
and
1
>
0,
maximum
if
3
>
0
and
1
<
0,
saddle
point
if
3
<
0
K
=
minimum
Check
Put the numbers back in. Both sides should match.
3
=
4
×
1
×
1
−
1
2
✓
2
=
1 × 0 − 2 × 1 × (−3)
3
✓
−1
=
1 × (−3) − 2 × 1 × 0
3
✓
−3
=
1
×
2
2
+
1
×
2
×
(−1)
+
1
×
(−1)
2
+
(−3)
×
2
+
0
×
(−1)
✓
3 > 0 and 1 > 0, so minimum ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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