ASSIGNMENT 3 - SOLUTIONS
1. Determine the nature of the critical points of
f (x, y) = exy ;
(a)
(b) f (x, y) = x2 + 2y 2 − xy ;
(c) f (x, y) = 3xey − x3 − e3y .
Solution. (a) We have (fx , fy ) = (yexy , xexy ). The critical points are the points where
(fx , fy ) = (0, 0), that is,
xy
ye
=0
so that (x, y) = (0, 0).
xexy = 0
The second order partial derivatives are
fxx (x, y) = y 2 exy ,
fxy (x, y) = exy + xyexy ,
fyy (x, y) = x2 exy .
Then
fxx (0, 0) = fyy (0, 0) = 0,
fxy (0, 0) = 1,
so that
D(0, 0) = fxx (0, 0) fyy (0, 0) − [fxy (0, 0)]2 = −1 < 0
which implies that (0, 0) is a saddle point.
(b) We have (fx , fy ) = (2x − y, 4y − x). The critical points are the points where
(fx , fy ) = (0, 0), that is,
2x − y = 0
or (x, y) = (0, 0).
4y − x = 0
The second order partial derivatives are
fxx (0, 0) = 2,
fxy (0, 0) = −1,
fyy (0, 0) = 4,
and
D(0, 0) = fxx (0, 0) fyy (0, 0) − [fxy (0, 0)]2 = 7 > 0.
Since fxx (0, 0) = 2 > 0 and D(0, 0) > 0 we deduce that (0, 0) is a local minimum
point1 for f .
(c) We have (fx , fy ) = (3ey − 3x2 , 3xey − 3e3y ). The critical points are obtained
by solving the system
3ey − 3x2 = 0
.
3xey − 3e3y = 0
Divide the second equation by 3ey to obtain x = e2y > 0. Substituting x = e2y in the
first equation we get ey = e4y , which implies y = 0. Then x = e2·0 = 1. Thus (1, 0)
is the only critical point.
The second order partial derivatives are
fxx (x, y) = −6x,
fxy (x, y) = 3ey ,
fyy (x, y) = 3xey − 9e3y .
1It is actually an absolute minimum since
x2 − xy + 2y 2 = (x −
1 2 7 2
y) + y ≥ 0 = f (0, 0).
2
4
1
2
ASSIGNMENT 3 - SOLUTIONS
Then
−6 3
D(1, 0) = 3 −6
= 27 > 0.
Since fxx (1, 0) = −6 < 0 and D(1, 0) > 0 we deduce that (1, 0) is a local maximum
point for f .
2. Let f (x, y) = x2 + 4y 2 − 4xy. Can you say whether the critical points are local
minima, maxima or saddle points?
Solution. We have (fx , fy ) = (2x − 4y, 8y − 4x). The critical points are obtained
by solving the system
2x − 4y = 0
so x = 2y.
8y − 4x = 0
The critical points are therefore (x, x2 ) with x ∈ R. The second order partial derivatives test is inconclusive since
x
D(x, ) = fxx (x, x/2) fyy (x, x/2) − [fxy (x, x/2)]2 = 2 · 8 − (−4)2 = 0.
2
However,
x
f (x, y) = (x − 2y)2 ≥ 0 = f (x, )
2
shows that all points (x, x2 ) with x ∈ R are absolute minimum points.
3. Find the points on the hyperboloid
z2 −
y 2 x2
−
=1
2
3
that are closest to the origin.
p
Solution. The distance from a point (x, y, z) to (0, 0, 0) is x2 + y 2 + z 2 . Let us
therefore minimize the function f (x, y, z) = x2 + y 2 + z 2 subject to the constraint
g(x, y, z) = 1, where
y 2 x2
g(x, y, z) = z 2 −
− .
2
3
The minimum points are found among the points on the hyperboloid where
∇f (x, y, z) = λ ∇g(x, y, z),
that is,
or

 (2x, 2y, 2z) = λ −2x , −y, 2z ,
3
 2 y 2 x2
z − 2 − 3 =
1.

x(1 + λ3 )
=0





 y(λ + 2)
=0






z(λ − 1)
2
=0
2
z 2 − y2 − x3 = 1.
From the first equation we deduce that either x = 0 or λ = −3.
If λ = −3 the system becomes

0
=0






−y
=0






−4z
z2 −
y2
2
−
=0
x2
3
= 1.
ASSIGNMENT 3 - SOLUTIONS
3
2
Hence y = z = 0 and the last equation above becomes − x3 = 1, which is a contradiction. Thus there are no solutions for λ = −3.
If x = 0 the system becomes

x
=0





 y(λ + 2) = 0
.

z(λ − 1) = 0




 2 y2
z − 2 = 1.
In this case either y = 0 or λ = −2. For λ = −2 we obtain z = 0 and then the last
2
equation above becomes − y2 = 1, which is again a contradiction. On the other hand,
if y = 0 we get z 2 = 1, which implies z = ±1. Then (x, y, z) = (0, 0, ±1). Both these
points have the distance to the origin equal to 1. Thus the points (x, y, z) = (0, 0, ±1)
are the closest to the origin.
4. Find the minimum value of x2 + y 2 + z 2 subject to the constraints x + y − z = 0
and x + 3y + z = 2. Give a geometrical interpretation of your answer and using it
explain why x2 + y 2 + z 2 has no maximum subject to these constraints.
Solution. Let
f (x, y, z) = x2 + y 2 + z 2 ,
g(x, y, z) = x + y − z,
h(x, y, z) = x + 3y + z.
The minimum points are found among the points with

 ∇f = λ∇g + µ∇h,
x + y − z = 0,

x + 3y + z = 2
that is,

 (2x, 2y, 2z) = λ (1, 1, −1) + µ (1, 3, 1),
x + y − z = 0,

x + 3y + z = 2.
Then
(∗)









x=
y=
λ+µ
2
λ+3µ
2
−λ+µ
2
z=





x + y − z = 0,



x + 3y + z = 2.
Substituting the values of x, y, z in the last two equalities above, we get
3λ + 3µ
3λ + 11µ
= 0 and x + 3y + z =
= 2.
x+y−z =
2
2
1
1
Hence λ = −µ and 3λ + 11µ = 4. Thus λ = − and µ = . Returning to (∗) we
2
2
1
1 1
1
obtain x = 0 and y = z = . We compute f (0, , ) = .
2
2 2
2
Let us now give a geometrical interpretation of our problem. We want to find the
extrema of x2 + y 2 + z 2 (that is, the square of the distance from (x, y, z) to the origin)
for (x, y, z) lying in the intersection of the two planes x + y − z = 0 and x + 3y + z = 2,
which is a line that we shall call L. There is obviously no maximum distance from
the origin to the given line but there is certainly a minimum distance. The square of
1
1 1
the minimum distance is and is obtained for (x, y, z) = (0, , ).
2
2 2