MAT3010 Problem Set 6: Suggested Solutions

Bin Liu

October 31, 2021

1. Answer: (a) The boundary of S can be found by making the inequality x2 + y 2 1 binding
so that the boundary is the set f(x; y) : x2 + y 2 = 1g. Since the inequality characterizing the set S—
x2 + y 2 1— is a weak one (it contains equality), S is a closed set. (Or S contains all its boundary
points so that it is closed.) Obviously, S is bounded.

Before we go to part (b) and part (c), let’s …rst establish the existence of the maximum point and min-
imum point. Notice that f (x; y) is a continuous function, and that the domain S is a closed and bounded
set— shown in part (a)— thus according to the extreme value theorem, the maximum and minimum points
exist.

(b) & (c) Let’s apply the procedure mentioned in the lecture. We will do part (b) and part (c)
together.

Step 1: Looking for category one candidates— stationary points. The …rst-order conditions are
fx0 (x; y) = 2x = 0 and fy0 (x; y) = 2y + 1 = 0. Thus, f has a single stationary point (x; y) = (0; 1=2),
which is in the constraint set (domain) S. The function value at (0; 1=2) is f (0; 1=2) = 5=4:

Step 2: Looking for category two candidates— the one on the boundary. The equation of the boundary
is x2 + y 2 = 1. From here we know that x2 = 1 y 2 . Substituting this into f (x; y) = x2 + y 2 + y 1
leads to a function just of y: g(y) = f (x; y) = 1 y 2 + y 2 + y 1 = y. [This is what we mean by
restricting to the boundary.] Our goal is to …nd the maximum/minimum point on the boundary. Notice
that x2 + y 2 = 1 implies that 1 y 1. Thus, the minimum point of g(y) is y = 1, and the
maximum point of g(y) is y = 1. Then, restricting to the boundary, the candidate for maximum point
is (0; 1), which leads to a function value f (0; 1) = 1; and the candidate for minimum point is (0; 1),
which leads to a function value f (0; 1) = 1.

Step 3: Comparing the function values at these candidates found. It is easy to see that f (0; 1=2)
is the smallest, while f (0; 1) is the largest. Therefore, we conclude that the maximum value for f (x; y)
is 1, achieved at (0; 1), and the minimum value for f (x; y) is 5=4, achieved at (0; 1=2).

2. Answer: First of all, the set S is a closed set. To see this, notice that both inequalities are
weak so that it is a closed set. (Another way to see this is …rst notice that the boundary of S is
f(x; y) : x2 + y 2 = 1 and x 0, or x2 + y 2 1 and x = 0g, obtained by equating one of the two
inequalities. Obviously, S contains all its boundary points.) It is also easy to see that S is bounded.
Now since f (x; y) is continuous and since the set it is de…ned on is closed and bounded set, f (x; y) has
maximum points and minimum points. Now let’s …nd the extreme points. We apply the same procedure
as before. But now I’m gonna skip some details.

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 Step 1: The FOCs are fx0 = 3x2 2x = 0 and fy0 = 2y = 0. The FOCs lead to two stationary
points (0; 0) and (2=3; 0), both in the set S. [footnote here.1 ] [Note: You need to check that the solution
to FOCs is indeed in the domain/constraint set.] The resulting function values are f (0; 0) = 3 and
f (2=3; 0) = 77=27:

Step 2: 2.1: Let’s …rst look at one piece of the boundary points: x2 + y 2 1 and x = 0. When x = 0,
2
y 1 and f (x; y) = 3 y 2 . Therefore, for this piece of boundary points, f (x; y) achieves the highest
value at (0; 0), which is f (0; 0) = 3; it achieves the lowest value at (0; 1), which is f (0; 1) = 2. 2.2:
Let’s turn to the second piece of the boundary points: x2 + y 2 = 1 and x 0. When x2 + y 2 = 1, we
have y 2 = 1 x2 , so that f (x; y) = 3 + x3 x2 (1 x2 ) = x3 + 2. Since x2 + y 2 = 1 and x 0, we
know that 0 x 1. Therefore, the highest function value on this piece of boundary points is obtained
at point (1; 0), which equals f (1; 0) = 3; the lowest function value on this piece is obtained at point
(0; 1), which is f (0; 1) = 2.

Step 3: Comparing the function values at all these candidates [footnote here2 ], we conclude that the
maximum value is 3, attained at (0; 0) and (1; 0); the minimum value is 2, attained at (0; 1) and (0; 1).
[In this question, we see that it is likely that the maximum/minimum point is not unique. But, of course,
the maximum/minimum value is unique.]

3. Answer: (a) FOCs: fx0 = 3x2 + 2y + 1 = 0 and fy0 = 2x + 2y = 0. From the second equation
we have y = x so that the …rst equation becomes 3x2 + 2x 1 = 0, which has two solutions x = 1
or 1=3. Therefore, there are two stationary points ( 1; 1) and (1=3; 1=3).

6x 2
The Hessian of f (x; y) is . It is easy to see that when x = 1=3, the Hessian is inde…nite
2 2
and that the determinant is nonzero. Thus, in this case, (1=3; 1=3) is a saddle point. When x = 1, it
is easy to see that the Hessian is positive de…nite, so that ( 1; 1) is a local minimum point. Now let’s
check whether ( 1; 1) is a global minimum point. In fact, f (2; 0) = 6 < 1 = f ( 1; 1). Thus, ( 1; 1)
is not a global minimum point.

(b) FOCs: fx0 = 2x = 0; fy0 = 2y = 0; and fz0 = 2z = 0. This leads to a single stationary
0 1
2 0 0
point (0; 0; 0). Notice that the Hessian matrix of f is @ 0 2 0 A for all (x; y; z), which is
0 0 2
negative de…nite. In particular, at point (0; 0; 0), the Hessian is negative de…nite so that (0; 0; 0) is a
local maximum point. Now notice that the Hessian matrix is negative de…nite for all (x; y; z), which
implies that f is (strictly) concave. This further implies that the stationary point we found, (0; 0; 0), is
also a maximum point.

4. Answer: (a) FOCs

@f @f @f
= 2x1 x2 + 2x3 = 0; = 2x2 x1 + x3 = 0; = 6x3 + 2x1 + x2 = 0:
@x1 @x2 @x3
1 Note that the stationary point (0; 0) is actually a boundary point of S, which is not an interior point. However, this

doesn’t matter at all. After all, we’ll always check the function values on the boundary, right? Therefore, even when the
stationary point solved is a boundary point, no need to delete it (or worry about this), just deal with it in the usual way.
2 Actually, you may have noticed that there are repetitions of the candidates in step 1, step 2.1, and step 2.2. But this

doesn’t matter, of course.

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This is a system of linear equations. Notice that the determinant of its coe¢ cient matrix is

2 1 2
1 2 1 = 4 6= 0:
2 1 6

Thus the system has only the trivial solution x = y = z = 0. Notice that we have shown in Q. 3.(c) of
Problem Set 5 that this function is strictly convex. Thus, (0; 0; 0) is a local minimum point, and in fact,
it is the (global) minimum point because f is convex.

(b) FOCs

@f @f @f @f
= 8x1 + 8x2 = 0; = 20 + 8x1 12x22 = 0; = 48 24x3 = 0; =6 2x4 = 0:
@x1 @x2 @x3 @x4
This yields (x1 ; x2 ; x3 ; x4 ) = ( 1; 1; 2; 3) or (5=3; 5=3; 2; 3) The Hessian of f is
0 1
8 8 0 0
B 8 24x 0 0 C
B 2 C:
@ 0 0 24 0 A
0 0 0 2

The leading principal minors: D1 = 8 < 0; D2 = 192x2 64; D3 = 24D2 ; D4 = 2D3 . Thus, the
Hessian is negative de…nite at (5=3; 5=3; 2; 3) and inde…nite at ( 1; 1; 2; 3). Thus, (5=3; 5=3; 2; 3) is a
local maximum point and ( 1; 1; 2; 3) is a saddle point since the determinant of the Hessian at this
point is nonzero.

5. Answer: (a) The claim does NOT generally hold. Consider the following single-variable function
f (x; y) = x(x 1)2 , de…ned on ( 1; +1). One can easily show that it has a unique local minimum
point x = 1. [The graph looks like the following.] However, x = 1 is not the minimum point for f ,
because lim f (x) = 1.
x! 1

[Comment: One can also easily construct examples for multivariable functions. What’s the problem
here? Well, the crucial point here for the failure of the claim is the existence of the minimum point!
In the example we constructed above, x = 1 is not the minimum point, because f does not have a
minimum point at all!

Now, on the other hand, in this question, if we further assume that the minimum point for f exists
in the domain S, then the claim must be true. Convince yourself that this is indeed the case. [A quick

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note: BTW, the example here also shows that f has a unique local maximum point in the domain,
but this point is NOT the maximum point for f !] Finally, this question emphasizes the importance of
existence of solutions before claiming a particular point is indeed the solution we need.]

(b) Proof: Suppose that f (x) is a strictly concave function de…ned on some convex set S Rn .
Our goal is to show that the maximum point for f , if exists, must be unique. Note …rst that if S is a
singleton (i.e., it contains only a single point), then the claim is trivially true (right?). To see that the
claim is true when S is not a singleton, suppose to the contrary that there exist x1 ; x2 2 S with x1 6= x2 ,
such that both x1 and x2 are maximum points for f . Thus, by de…nition, f (x1 ) = f (x2 ) f (x), for all
x 2 S. However, for x3 = 21 x1 + 12 x2 2 S (since S is a convex set), we have

1 1
f (x3 ) > f (x1 ) + f (x2 ) = f (x1 ) f (x), for all x 2 S:
| 2 {z 2 } | {z }
x1 is m ax p oint
f is strictly concave

But f (x3 ) > f (x) for all x 2 S is a contradiction, since f (x3 ) > f (x3 ) is impossible.

BTW, what should be the corresponding conclusion for a convex function?

[Comment: Note that the question itself does NOT claim the existence of maximum points. It
rather claims the uniqueness, given that the maximum point exists. Convince yourself by thinking of
an example where f is strictly concave but it does not have a maximum point!]

6. Answer: FOCs:
@f @f
= r2 2x = 0; = 3s2 16y = 0:
@x @y
The solution is x = r2 =2 and y = 3s2 =16. Notice that the function is concave in (x; y) because r2 x,
3s2 y; x2 ; and 8y 2 are all concave in (x; y) and the sum of concave functions is still concave. Therefore,
as the stationary point for a concave function, (x (r; s); y (r; s)) = (r2 =2; 3s2 =16) is a maximum point.

The value function

r4 9
f (r; s) = f (x (r; s); y (r; s); r; s) = + s4 :
4 32
@f @f 9s3
So @r = r3 ; @s = 8 :

This is the direct way. Let’s check the envelope theorem.

@f @f
= (x; y; r; s)j(x;y)=(x ;y ) = 2rx (r; s) = r3
@r @r
@f @f 9s3
= (x; y; r; s)j(x;y)=(x ;y ) = 6sy (r; s) = :
@s @s 8
It gives the same answer.

7. Answer: (a) For t > 0, f (tx; ty) = (tx)2 (ty) + (tx)3 = t3 (x2 y + x3 ) = t3 f (x; y), so f (x; y)
is homogeneous of degree 3. For the Cobb-Douglas function f (x) = Ax1 1 x2 2 xnn , since f (tx) =
1+ 2+ + n 1+ 2+ + n
At x1 x2
1 2
xn = t
n
f (x), f (x) is homogeneous of degree 1 + 2 + + n.

(b) Since f (tx) = tk f (x), taking partial derivative w.r.t. xi on both sides of the equation leads to
(note that the LHS uses chain rules)
fi0 (tx) t = tk fi0 (x);

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so fi0 (tx) = tk 1 0
fi (x), which means that fi0 (x) is homogeneous of degree k 1.

(c) Suppose that the level curve is f(x; y) 2 D : f (x; y) = cg, where c is a constant. At point
(x; y) with f (x; y) = c, the slope of the tangent line at (x; y) to f (x; y) = c can be found by implicit
di¤erentiation: the slope is f10 (x; y)=f20 (x; y). We show that this slope is homogeneous of degree 0. In
fact, at point (tx; ty), the slope is f10 (tx; ty)=f20 (tx; ty). We know from part (b) that f10 and f20 are
homogeneous of degree k 1. Thus,
f10 (tx; ty) tk 1 0
f1 (x; y) f10 (x; y)
= = ;
f20 (tx; ty) tk 1 f 0 (x; y)
2 f20 (x; y)
so f10 (x; y)=f20 (x; y) is homogeneous of degree 0.

(d) The “only if” direction: Suppose that f (x) is homogeneous of degree k, then by de…nition,

f (tx) = tk f (x); (1)

for any t > 0 and any x 2 D. Now …xing x = (x1 ; x2 ; : : : ; xn ) and di¤erentiating w.r.t. t on both sides
of (1) leads to
n
X
fi0 (tx)xi = ktk 1 f (x); for any t > 0.
i=1
Pn
Evaluating the above equation at t = 1 leads to i=1 xi fi0 (x) = kf (x).

The “if” direction: Suppose that

n
X
xi fi0 (x) = kf (x); for any x 2 D: (2)
i=1

Our goal is to show that f (tx) = tk f (x), for any x 2 D and any t > 0. This is equivalent to showing
that
t k f (tx) = f (x); for any x 2 D and any t 2 (0; +1):
To this end, …x x = (x1 ; x2 ; : : : ; xn ) and de…ne a single-variable function
k
g(t) = t f (tx) f (x); t 2 (0; +1):

If we can show that g(t) = 0 for any t > 0, then the proof is done. Note that g(1) = f (x) f (x) = 0.
Therefore, the proof boils down to showing that g 0 (t) = 0 for any t > 0, because this implies that g( ) is
a constant over t 2 (0; +1) and this constant must be g(1) = 0.

Now let’s check that g 0 (t) = 0 for any t > 0. In fact, chain rule implies that
n
X
g 0 (t) = kt k 1
f (tx) + t k
xi fi0 (tx): (3)
i=1
Pn
Since (2) holds
Pn for any x 2 D, it also holds for tx. Replacing x with tx in (2) gives i=1 txi fi0 (tx) =
0 1
kf (tx), so i=1 xi fi (tx) = t kf (tx). Thus, the second term in the RHS of (3) can be rewritten,
leading to
n
X
g 0 (t) = kt k 1
f (tx) + t k
xi fi0 (tx) = kt k 1
f (tx) + t k
t 1
kf (tx) = 0:
i=1

This completes the proof.