Math 21a
Maxima, minima, and saddle points
Fall 2016
1 Find the critical points of each of the following functions.
(a) f (x, y) = x2 + y 2
(b) f (x, y) = −x2 − y 2
(c) f (x, y) = x2 − y 2
(d) f (x, y) = x2
Do the critical points above correspond to maxima, minima, or neither (saddle points)?
Solution.
(a) ∇f (x, y) = h2x, 2yi, so ∇f (x, y) = 0 iff (x, y) = (0, 0). Since this is the graph of an
elliptic paraboloid, this critical point (0, 0) is a minimum.
(b) ∇f (x, y) = h−2x, −2yi, so ∇f (x, y) = 0 iff (x, y) = (0, 0). This is the graph of an
upside-down elliptic paraboloid, so this critical point is a maximum.
(c) ∇f (x, y) = h2x, −2yi, so ∇f (x, y) = 0 iff (x, y) = (0, 0). This is the graph of a hyperbolic
paraboloid, and this critical point is a saddle point.
(d) ∇f (x, y) = h2x, 0i, so ∇f (x, y) = 0 iff x = 0, and y can be anything. Since this is the
graph of a parabolic cylinder (a parabola in the xz-plane, shifted along the y-direction),
all these critical points are (non-strict) minima.
2 Check that the second derivative test works for the functions in the previous problem.
Solution.
(a) At the critical point (0, 0), we have
2 0 =4>0
D = 0 2
and fxx = 2 > 0, so this is indeed a minimum by the second derivative test.
(b) At the critical point (0, 0), we have
−2 0 =4>0
D = 0 −2
and fxx = −2 < 0, so this is indeed a maximum by the second derivative test.
(c) At the critical point (0, 0), we have
2 0 = −4 < 0
D = 0 −2
, so this is indeed a saddle point by the second derivative test.
(d) At the critical point (0, y), we have
2 0 =0
D = 0 0
, so the second derivative test is inconclusive (even though we know that these critical
points are nonstrict minima).
3 Find and classify all the critical points of the following functions.
(a) f (x, y) = (x2 − 1)(y 2 − 4)
(b) f (x, y) = ey (y 2 − x2 )
(c) f (x, y) =
x3
3
−x−
y3
3
+y
(d) f (x, y) = x3 − 3xy 2 (this is called the “monkey saddle”)
(e) f (x, y) = x4 + y 4
Solution.
(a) We have ∇f (x, y) = h2x(y 2 − 4), 2y(x2 − 1)i, so there are five critical points: (0, 0),
(1, 2), (1, −2), (−1, 2), and (−1, −2). The Hessian matrix is
2
2(y − 4)
4xy
fxx fxy
=
,
fyx fyy
4xy
2(x2 − 1)
so D = 4(x2 − 1)(y 2 − 4) − 16x2 y 2 . We need to evaluate the signs of D and fxx at the
critical points. We summarize our calculations in the following table:
point
(0, 0)
(1, 2)
(1, −2)
(−1, 2)
(−1, −2)
D
16
−64
−64
−64
−64
fxx
−8
?
?
?
?
classification
maximum
saddle point
saddle point
saddle point
saddle point
value
4
0
0
0
0
(b) We have ∇f (x, y) = h−2xey , 2yey + (y 2 − x2 )ey i, so there are two critical points (0, 0)
and (0, −2). The Hessian matrix is
fxx fxy
−2ey
−2xey
=
,
fyx fyy
−2xey (2 + 4y + y 2 − x2 )ey
so D = −2(x2 + y 2 + 4y + 2)e2y . We need to evaluate the signs of D and fxx at the
critical points. We summarize our calculations in the following table:
point
(0, 0)
(0, −2)
D
−4
4e−4
fxx
?
−2e−2
classification
saddle point
maximum
value
0
4e−2
(c) We have ∇f (x, y) = hx2 − 1, −y 2 + 1i, so there are four critical points (1, 1), (1, −1),
(−1, 1), and (−1, −1). The Hessian matrix is
fxx fxy
2x 0
=
,
fyx fyy
0 −2y
so D = −4xy. We need to evaluate the signs of D and fxx at the critical points. We
summarize our calculations in the following table:
point
D
(1, 1)
-4
(1, −1)
4
(−1, 1)
4
(−1, −1) −4
fxx
?
2
−2
?
classification
saddle point
minimum
maximum
saddle point
value
0
− 34
4
3
0
(d) We have ∇f (x, y) = h3x2 − 3y 2 , −6xyi, so there is a single critical point (0, 0). The
Hessian matrix is
fxx fxy
6x −6y
=
,
fyx fyy
−6y −6x
so D = 36(y 2 −x2 ). At (0, 0), we have D = 0, so the second derivative test is inconclusive.
It turns out that (0, 0) is a saddle point. If we slice the graph of f (x, y) with the vertical
plane y = kx, we get the graph of f (x, kx) = (1 − 3k 2 )x3 , which is a flex point at the
origin.
(e) We have ∇f (x, y) = h4x3 , 4y 3 i, so the unique critical point is (0, 0). The Hessian matrix
is
12x2
0
fxx fxy
=
,
fyx fyy
0
12y 2
so D = 144x2 y 2 . At (0, 0), we have D = 0, so the second derivative is again inconclusive.
But we see that the graph of f (x, y) is a “deformed” elliptic paraboloid, so (0, 0) is in
fact a minimum. Alternatively, we see that x4 + y 4 = (x2 )2 + (y 2 )2 is a sum of squares
and so is always non-negative. At (0, 0), we have f (0, 0) = 0, so (0, 0) is a (global)
minimum.
There is a reason why the second derivative test failed in the last two examples. Essentially, the second derivative test works by finding a quadratic approximation to the function
f , but in parts (d) and (e) we have homogeneous cubic and quartic polynomials, which are
“invisible” to the eyes of quadratic approximation. They carry higher-order information, so
we’ll need a “higher derivative test” to study the nature of their critical points.
4 Consider a physical system of three springs attached in series whose two ends are fixed a unit distance
apart. Let x be the length of the first spring, and let
y−x be the length of the second spring. By Hooke’s law,
the potential energy of a spring is proportional to the
square of its displacement, so the total energy is given
by f (x, y) = x2 + (y − x)2 + (1 − y)2 . Find the unique
equilibrium (critical point) of this system. Is it stable
(a local minimum) or unstable (not a local minimum)?
Solution. We have
∇f (x, y) = h4x − 2y, −2x + 4y − 2i,
and the unique critical point of f is ( 31 , 23 ). This is a local minimum because at ( 31 , 23 ), we
have D = 12 > 0 and fxx = 4 < 0. This answer agrees with our physical intuition.
5 The centroid, aka the center of gravity, of a triangle is the point that minimizes the sum
of the squared distances to each vertex. Find the centroid of the triangle with vertices at
(0, 0), (4, 4), and (5, 2).
Solution. We want to minimize
f (x, y) = x2 + y 2 + (x − 4)2 + (y − 4)2 + (x − 5)2 + (y − 2)2 .
We have
∇f (x, y) = h6(x − 3), 6(y − 2)i,
so the unique critical point is (x, y) = (3, 2). At this point, we have fxx = 6, fxy = 0, an
fyy = 6, so the discriminant is D = 36. Since D > 0 and fxx > 0, this is a local minimum, i.e.,
the centroid of the triangle is (3, 2) (which happens to be the average of the three vertices).
Moreover, lim(x,y)→∞ f (x, y) = ∞, the point (3, 2) is in fact a global minimum.