ECOS2903 Final Exam Answers S1 2022

Total: 33 points

Q1. (8 points) Consider the function

Version 1

∫ : {（x← y）Ix > 0← y > 0} - R s.t. ∫（x← y）=（x4 +y）2

Version 2

∫ : {（x← y）Ix > 0← y > 0} - R s.t. ∫（x← y）= (x0．5 +y)0．5

(a) Is ∫ concave? Is ∫ convex? (4 points)

(b) Is ∫ quasi-concave? Is ∫ quasi-convex? (4 points)

Answer:

Marking guideline: 2 pts each for concavity, convexity, quasi-concavity, and quasi-convexity.

Version 1:

H =  l  8（7x68x(+)33x2y） 82(x)3   ]

IHI = 8（12x6 + 3x2y）> 0, 8（7x6 + 3x2y）> 0, positive definite. ∫ is (strictly) convex and, therefore, quasi-convex.

Students can prove that ∫ is not concave by showing that they can find two points（x1 ← y1）and  （x2 ← y2）in the domain of ∫ and & e（0← 1）such that（x&←y&）=（&x1 +（1 _ &）x2 ← &y1 +（1 _ &）y2）is also in the domain but f（x&←y&）> &f（x1）+（1 _ &）f（x2）.  To get full marks, students must either (a) explicitly state the values of（x1 ← y1）,（x2 ← y2）, &, f（x1 ← y1）← f（x2 ← y2）, and f（x&←y&）, OR (b) show an accurate plot of the function using software and draw on the plot to illustrate where （x1 ← y1）,（x2 ← y2）,（x&←y&）are and why they violate concavity.

There are two methods to prove that f is not quasi-concave:

Method 1:  Show that it violates the necessary condition in the bordered Hessian test.   The bordered Hessian is

B = {┌ 8x3 ╱x4(0) +y← 87(x)x(3) 3x2(+y) ） 2 ╱8x(x4)y←

|   2 ╱x4 +y← 8x3                       2

┐

ì

|

When x > 0 and y > 0, its order-2 leading principal minor is D2  = _8x6 ╱x4 +y←2 ← 0 and its order-3 leading principal minor is D3  = _96x14 _ 288x10y _ 288x6y2 _ 96x2y3  ← 0. However, the necessary condition for quasi-concavity requires that D2  < 0 and D3  2 0 (see end of lecture 8). Therefore, this necessary condition for quasi-concavity is violated and f is not quasi-concave.

Method 2: Find three points in the domain that violate quasi-concavity. You can either provide a plot from Mathematica and annotate on it to show the violation or explicitly give the numerical coordinates of the three points. For example, from the contour plot below we can see that f（A）= f（B）because they are on the same indifference curve.  Point C = &A +（1 _ &）B for some & e （0← 1）because it’s on the segment AB, but f（C）← min {f（A）← f（B）} because C is on a lower  indifference curve. This violates the definition for quasi-concavity (see PS2 Q1).

Alternatively, you can give explicit coordinates without attaching any graph, e.g., let（x1 ← y1）= （0．2 ← 0．7）,（x2 ← y2）=（0．8 ← 0．3）, & = 0．5 so that（x&←y&）=（0．5 ← 0．5）. f（x1 ← y1）= 0．4922, f（x2 ← y2）= 0．5035, f（x&←y&）= 0．3164 ← min{f（x1 ← y1）← f（x2 ← y2）}.

Note that it’s incorrect to state that “all monotonic functions are both quasi-concave and quasi- convex”. This statement is true if fis a single-variable function and not true iff is multivariate. For example, Cobb-Douglas functions on R

Version 2:

┌  _ H = {

|

0．125

_

x0．5（x0．5 +y）1．5 0．25

_

（x0．5 +y）1．5

┐

ì

|

IHI = x1．5(0)x(0)0(3)5(1)）2   > 0, _x（x0(0)．(．)5(0) 1．5  _ x1．5（x(0)0(．)y）0．5 ← 0, negative definite. f is (strictly) con-

cave and, therefore, quasi-concave.

f is not quasi-convex and, therefore, not convex. To show this, notice from the contour plot that one can find two points on the same level set off , and their middle point is on a higher level set. This shows that it’s possible to find（x1 ← y1）and（x2 ← y2）in the domain off and & e（0← 1）such that（x&←y&）=（&x1 +（1 _ &）x2 ← &y1 +（1 _ &）y2）, but f（x&←y&）> max {f（x1 ← y1）← f（x2 ← y2）}. Equivalently, students can show that it’s possible to find ce R such that f（x1）< candf（x2）< c but f（x&）> c.  This leads to a violation to the definition of a quasi-convex function.  To get full marks, students must explicitly state the values of（x1 ← y1）,（x2 ← y2）, &, f（x1 ← y1）← f（x2 ← y2）, and f（x&←y&）.

Q2. (8 points) A firm uses labour (L) and capital (K) to produce an output (Y). The production technology is described by the production function Y（L← K）= ln（LK）. Quantities of labour and capital are non-negative, i.e., L 2 0 and K 2 0. The price of the output is py, the price of labour is pL, and the price of capital is pK. Can you help the firm find a bundle（L* ← K*）that maximises its profit? To get full marks you need to check both the necessary and sufficient

conditions. If（L* ← K*）exists, please state what it is and calculate dY K* ）at py = version.

If you cannot find a profit-maximising bundle, please explain why it doesn’t exist.

version 1: py = 10, pL = 2, pK = 1.

version 2: py = 20, pL = 4, pK = 5.

version 3: py = 15, pL = 2, pK = 3.

Answer:

The profitfunction is

π（L← K）= py ln（LK）_ pLL _ pKK

The two constraints L 2 0 and K 2 0 are never binding because ln0 is undefined (lim lnx -

the solution (if there is any) satisfies L > 0 and K > 0.

max π（L← K）= py（lnL +lnK）_ pLL _ pKK

L ←K

Necessary condition:

2(2)L(π) = 0 ÷ L(py) = pL ÷ L* =pL(py) > 0

2(2)K(π) = 0 ÷ K(py) = pK ÷ K* =p K(py) > 0

Sufficient condition:

π（L← K）is a concave function. Its Hessian matrix is

H = - _2(y) ┐

whose leading principals are _ This proves that H is negative definite and π is strictly concave.

Therefore, the critical point（L* ← K*）= ╱ p(p)L(y) ← p K(py) 、is indeed the solution to this profit-maximising

problem. In version 1,（L* ← K*）=（5← 10）. In version 2,（L* ← K*）=（5← 4）. In version 3,（L* ← K*）=

（7．5 ← 5）.

Note: The Extreme Value Theorem is not applicable to this problem because the feasibility set {（L← K）IL 2 0← K 2 0} is not bounded.

Y（L* ← K*）= ln ╱p(p)L(y) .p K(py)、= 2 ln py _ln pL _ln pK, so

dY（L* ← K*） d py

2

=

py

= = 0．2 in version 1

= = 0．1 in version 2

= = 0．1333．．．in version 3

Note that we cannot directly use the envelope theorem here because the envelope theorem can

only help us calculate d π p(*)y(←)K*）, which is different from dY K*）.

Marking guideline:

2 pts: Writing down the correct maximisation problem.

2 pts: Correctly stating and solving the necessary conditions

2 pts: Correctly stating and proving the sufficient condition

2 pts: Correct calculation of dY K*）. Given 1 pt if student used the envelopment theorem to calculate d π p(*)y(←)K* ）instead of dY y(←)K* ）and their calculation of d π K* ）is correct (= ln L* + lnK*

= 3.912 in version 1, 2.996 in version 2, and 3.624 in version 3).

Q3. (8 points) Suppose that Ann’s utility is a function of three goods,x, y, and z, based on the  function U（x← y← z）= x2 +Axy +y2 +z2 where x 2 0, y 2 0,and z 2 0. Ann’s total expenditure  on these three goods must be weakly lower than M dollars.  The price for x is px, the price  for y is py, and the price for z is pz . Does Ann have a utility-maximising bundle（x* ← y* ← z*） that satisfies all the constraints?  If so, state what the solution is and explain the steps you  take to find the solution. If not, please explain why a solution does not exist. (Hint: use your  intuition to reduce the number of constraints and/or variables.)

Version 1: A = 3, M = 10, px = py = 1, pz = 1.

Version 2: A = 4, M = 15, px = py = 2, pz = 3.

Answer:

There are four constraints: (1) pxx + pyy + pzz < M, (2) x 2 0, (3) y 2 0, and (4) z 2 0. Below, I argue that (1) and (4) must bind. (2) and (3) cannot be both binding.

Constraint (1) must bind (pxx + pyy + pzz must be equal to M): Because U is strictly increasing  in x← y← z, we must have pxx + pyy + pzz = M at the optimum.  Prove by contradiction:  If（x← y← z） is the solution to the maximisation problem and pxx + pyy + pzz ← M, then we can find a small  enough ε > 0 such that px（x + ε ) +pyy +pzz < M and U（x + ε ← y← z）> U（x← y← z）because U strictly  increases in x, keeping y and z fixed. This shows that（x← y← z）cannot be a solution.

Constraint (4) must bind (z must be 0): Notice that z is weakly more expensive than x in both  versions.  However, z has a smaller contribution than x in U, which makes it a strictly inferior  investment.  Proof:  Consider a bundle（x← y← z）where z > 0.  Consider another bundle（x +z ← y← 0）.  If（x← y← z）is affordable then（x +z ← y← 0）is also affordable because x is weakly less expensive than z. Moreover, for any x 2 0 and y 2 0, U（x +z ← y← 0）=（x +z）2 +A（x +z）y +y2 > x2 +Axy +y2 + z2  = U（x← y← z）.  Therefore, U cannot be maximised when z > 0,because U（x← y← z）← U（x +z ← y← 0） whenever z > 0.

The fact that Constraints (1) and (4) bind imply that (2) and (3) cannot both bind.  Because M > 0, either x or y must be strictly positive.

Therefore, we can simplify the question to the following:

x(m) U（x← y）= x2 +Axy +y2  s.t. pxx + pyy = M ← _x < 0← _y < 0

The Lagrangian function is

ψ（x← y← λ1 ← λ2 ← λ3）= x2 +Axy +y2 _ λ1（pxx + pyy _ M）+λ2x +λ3y

The necessary conditions for a solution are

「x|   2x +Ay _ λ1px+ λ2     = 0

「y| Ax + 2y _ λ1py + λ3     = 0

λ1 > 0← pxx + pyy = M

λ2  > 0← x > 0← λ2x = 0

λ3  > 0← y > 0← λ3y = 0

We discuss only three cases:

Case A: pxx +pyy = M, x > 0 andy > 0. This implies that λ2 = λ3 = 0, and a solution candidate must satisfy

「x|    2x +Ay = λ1px

「y| Ax + 2y = λ1py

「λ1| pxx + pyy = M

These two equationsimply that

x =

y =

λ1 =

M（2px _Apy）

2 ╱px(2) _Apxpy + py(2)←

= 5 version 1

= 5 version 1

= 25 version 1

or 3．75 version 2

or 3．75 version 2

or 11．25 version 2

which satisfy the conditions that x > 0, y > 0 and λ1  > 0.  Therefore,（x← y）=（5← 5）is a valid solution candidate.

Case B: pxx + pyy = M, x > 0 and y = 0. x > 0 implies that λ2  = 0. Plugging λ2  = y = 0 into 「x| and「y| we get

「x| x =

「y| x =

These two equations require that = , which is never true when px = py , A > 2, and

λ3 2 0. Therefore, Case B does not give us any solution candidate.

Case C: pxx + pyy = M, x = 0 and y > 0. y > 0 implies that λ3  = 0. Plugging λ3  = x = 0 into 「x| and「y| we get

「x|   2x +Ay _ λ1px+ λ2     = 0

「y| Ax + 2y _ λ1py + λ3     = 0

These two equations require that

「x| y

「y| y

λ1px__λ2

A

λ1px _ λ2

=

A

λ1py

=

2

= , which is never true when px = py , A > 2, and

λ2 2 0. Therefore, Case C does not give us any solution candidate.

To conclude, the only solution candidate that satisfies all the necessary conditions is（x* ← y*）= （5← 5）in version 1 and（x* ← y*）=（3．75 ← 3．75）or ← in version 2.

Note: It’s not wrong to skip the discussion of constrains z = 0 or pxx + pyy = M.  However,  if the student did not first prove that these constraints are binding, they must include more case discussions when they solve the necessary conditions.  The result is the same even with more case discussions: the only valid solution candidate is（x* ← y*）=（5← 5）in version 1 and（x* ← y*）= （3．75 ← 3．75）or ← in version 2.

Students must complete either (A) or (B) below to prove that（x* ← y* ← 0）is the real maximiser of U subject to the constraints:

(A) Check sufficient conditions:

Check U: When z = 0, U（x← y）= x2 +Axy + y2  is not concave, but it is quasi-concave with a non-zero gradient at（x* ← y*）. To see why, note that its bordered Hessian is

┌ 0        2x +Ay Ax + 2y ┐

B =  {  2x +Ay 2 A ì

| Ax + 2y A 2      |

When x and y are non-negative and not both zero, the bordered Hessian matrix’s order-2 leading

principal minor is _（2x +Ay）2 ← 0, and its order-3 leading principal minor is 2 ╱A2 _ 4← ╱x2 +Axy + y2←> 0 in both versions.  This satisfies the sufficient conditions for quasi-concave functions, so U（x← y）

is quasi-concave. Its gradient vector is

▽U = lAx(2x)2y(Ay) ] l 0(0) ]

at our solution candidate（x* ← y*）.

Check the constraint function: g（x← y）= pxx + pyy is a linear function, which implies that it is convex and therefore quasi-convex.

Therefore, the solution candidate（x* ← y*）=（5← 5）in version  1 and（x* ← y*）=（3．75 ← 3．75）or ← maximisation problem and is indeed the real solution.

Alternatively,

(B) Apply the Extreme Value Theorem:

The feasible set D = {（x← y← z）| pxx + pyy + pzz 三 M← x > 0← y > 0← z > 0}is a compact tetrahed- ron and the objective function U（x← y← z）is continuous. Therefore, the maximisation problem must have at least one solution, and this solution must satisfy all the necessary conditions. Because our solution candidate（x* ← y* ← 0）is the only point that satisfy all necessary conditions, it must be the real solution to our maximisation problem.

Marking guideline:

1 pts: Correctly states the Lagrangian function (0.5 pt) and its associated necessary conditions given the student’s version of the Lagrangian function (0.5 pt).

5 pts: Correctly solve the necessary conditions to find the solution candidate（x* ← y*）.  Partial marks are given on a case-by-case basis.

2 pts: (A) Correctly stating and solving the sufficient conditions OR (B) correctly applying the Extreme Value Theorem to prove that the solution candidate from the necessary conditions is the real solution.

Q4. (9 points) Consider the following minimisation problem

ex + yA s.t. xy 2 A