8/11/2019 Beem 103 Test 10 Sol

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BEEM103

January 2010OPTIMIZATION TECHNIQUES FOR ECONOMISTS

solutions

Part A (You can gain no more than 55 markson this part.)

Problem 1 (10 marks) Simplify

1

8

px2y1

x3 2

3

1

x2p

y and

x= 3p

x2

x7

3

Solution 1

1

8px2y1

x3 2

3

1

x2p

y =

3

24

1

x2p

y 16

24

1

x2p

y = 13

24

1

x2p

y

x= 3px2

x5

3

= x=x

2

3

x7

3

=x

1

3

x7

3

=x1

3 7

3 =x6

3 =x2 = 1

x2

Problem 2 (10 marks) Solvex3lnx =x2+lnx

Solution 2

ln x3lnx = (3 ln x) ln xln x2+lnx = (2 + lnx) ln x

(3 ln x) ln x = (2 + lnx) ln x(1 + 2 ln x) ln x = 0

Solutions: ln x= 0, ie. x= 1, orln x= 12

orx = e1

2 =p

e

Problem 3 (10 marks) Consider the function

y (x) =xex2

2

i) Calculate and draw a sign diagram for the rst derivative. Where is the functionincreasing or decreasing? Are there any peaks or troughs? Is the function quasi concave

on the interval of positive numbers x > 0? Does the function have a global maximumover the positive numbers x >0?

ii) Calculate and draw a sign diagram for the second derivative. Where is the functionconvex or concave? Are there any inection points?

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Solution 3

-5 -4 -3 -2 -1 1 2 3 4 5

-0.4

-0.2

0.2

0.4

x

y

a) Using the product and chain rule we obtain

y0 = (x)0 ex2

2 +x

ex2

2

0=

1 x2

e

x2

2

Since ex2

2 ia always positive there are critical points at1 and+1. Because 1 x2 ispositive inside (1; 1) and negative outside [1; 1] ; the function is increasing inside anddecreasing outside the interval [1; +1] : Thus+1is a peak,1 a trough. On the set ofpositive numbers the function is increasing until it reaches a global maximum at +1 andis then decreasing. In particular, it is quasiconcave on the positive numbers.

b)

y00 = (2x) ex2

2 2x 1 x2 ex22= 2x 2 x

2

ex

2

2

The second derivative is zero at x= 0;p2. It is non-negative (and the function henceconvex) on the intervals

p2; 0 and p2; +1. It is non-positive (and the functionhence concave) on the remaining intervals.

Problem 4 (10 marks) For the function

y= x4 2x2

nd the (global) maxima and minima a) on the interval [1; 1] and b) on the interval[0; 2] :

Solution 4

y0 = 4x3 4xThe derivative is zero when x = 0orx = 1. We have y (1) = 1,y (0) = 0.

On the interval [1; 1] the maximum is hence at x = 0 while x = 1 andx =1 arethe two minima. Sincey (2) = 8 the maximum on[0; 2]is atx = 2while the minimum isatx = 1.

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This is illustrated by the graph of the function:

-2 -1 1 2

2

4

6

8

x

y

Problem 5 (10 marks) Find the equation of the tangent plane of

z (x; y) = ln (5x +y) + ln (10x+y)

at the point(x; y; z) = (2; 3; z (2; 3)).

Solution 5

@z

@x =

5

5x+y+

10

10x+y@z

@x j(2;3)=

5

10 + 3+

10

20 + 3=

5

13+

10

23=

245

299@z

@y =

1

5x+y+

1

10x+y@z

@y j(2;3)=

1

10 + 3+

1

20 + 3=

1

13+

1

23=

36

299

z (2; 3) = ln 13 + ln 23

The equation for the tangent is hence

z =

5

13+

10

23

(x 2) +

1

13+

1

23

(y 3) ln13 ln23

=

5

13+

10

23

x+

1

13+

1

23

y ln13 ln17 2

5

13+

10

23

3

1

13+

1

23

= 245

299x +

36

299y ln13 ln17 2

Problem 6The only grocery store in a small rural community carries two brands offrozen apple juice, a local brand that it obtains at the cost of 30 cent per can and a

well-known national brand that it obtains at a cost of 40 cent a can. The grocer estimatesthat if the local brand is sold for x cents per can and the national brand for y cents percan, approximately(70 5x+ 4y)cans of the local brand and (80 + 6x 7y)cans of thenational brand will be sold each day. How should the grocer price each brand to maximizethe prot from the sale of the juice?

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Solution 6 The prot function is

(x; y) = (x 30)(70 5x+ 4y) + (y 40) (80 + 6x 7y)We have (using the product rule)

@

@x = 1 (70 5x + 4y) + (x 30) (5) + (y 40) 6 = 10y 10x 20@

@y = (x 30) 4 + 1 (80 + 6x 7y) + (y 40) (7) = 10x 14y+ 240

10x 14y+ 240 and obtain the system of simultaneous equations10x+ 10y 20 = 010x 14y+ 240 = 0

Addition yields4y+ 220 = 0 ory = 55: From the rst equation x = y 2 = 53: Thus(x; y) = (53; 55) is the unique critical point of the function. The Hessian matrix is

H=

" @2@x2

@2@y@x

@2@x@y

@2@y2

#=

10 1010 14

We have @2

@x2 = 10< 0 and

det H = (10) (14) 10 10 = 40> 0= 100 140 100 100 = 100 (140 100)> 0

so the function is strictly concave and the critical point hence a (global) prot maximum.

Problem 7 (10 marks) Find a solution to the dierential equation

dxdt

=t4x2

Solution 7

x2dx = t4dtZ x2dx =

Z t4dt

x1 = 15

t5 +C

x =

5

t5

+CTest: dx

dt =

5 (t5 +C)2 5t4

= 25 t

4

(t5+C)2,t4x2 =t4 25

(t5+C)2

Problem 8 (10 marks) Solve the problem

minx(t)

Z 10

1 +x +x2 + _x+ _x2

ertdt

subject to the restrictions x (0) = 0,x (1) = 1.

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Solution 8

F(t;x; _x) =

1 +x+x2 + _x+ _x2

e3

2t

@F

@x = (1 + 2x) e

3

2t

@F@_x

= (1 + 2 _x) e3

2 t

d

dt

@F

@_x =

3

2+ 3 _x + 2x

e3

2t

The Euler equation becomes

(1 + 2x) e3

2t =

3

2+ 3 _x+ 2x

e3

2t

1 + 2x = 3

2+ 3 _x+ 2x

12

= 2x+ 3 _x 2x

14

= x +3

2_x x

A special solution to this dierential equation isx (t) = 14 , but this would not satisfy theboundary conditions. The characteristic polynomial for the homogeneous DE x+ 3

2_xx=

0 is

0 =2 +3

2 1

which has the roots 12 ;2. The general solution to the Euler equation is hence

x (t) = 14

+Ae1

2t +Be2t

The boundary conditions are now

0 = 14

+A +B

1 = 14

+Ae1

2 +Be2

which have the solution

A= e2 54e2 4e 12

; B= e1

2 54e2 4e 12

Our unique candidate for an optimum is thus

x (t) = 14

+ e2 54e2 4e 12

e1

2t e

1

2 54e2 4e 12

e2t

and this is indeed a minimum becauseF is concave in both arguments.

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Part B (You can gain no more than 15 markson this part.)

Problem 9 (15 marks) Determine the optimal prot for a rm with the productionfunction

Q=A ln K+B ln L

(A; B >0) operating in perfectly competitive input- and output markets and facing theoutput priceP >0; the interest rate r >0 and the wage rate w.

Solution 9

=P Q rK wL

@

@K = AP

1

K r= 0

@

@K = BP

1

Lw= 0

Division yields

AL

BK =

r

w

K = A

B

w

rL

Plugging this into the rst equation above yields

APB

A

r

w

1

L=r , P B

wL = 1 , L= BP

w

and symmetrically we obtain

K=

AP

rThe solution is economically meaningful because both KandL are positive. To see thatwe indeed have a maximum we calculate the Hessian matrix is

H=

AP K2 00 BP L2

which has a negative top-left entry and a positive determinant

det H=ABP2K2L2 >0

The prot function is thus convex and the critical point a global maximum.

Problem 10 (15 marks) For the point (1;3)nd the point (x; y)that minimizes thedistance q

(x a)2 + (y b)2

subject to the constraints

1 x 1x 1 y x+ 1:

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Solution 10 A graph indicates that only the constraint y = x 1 is binding in theoptimum. We start hence with the Lagrangian

L = (x 1)2 (y+ 3)2 + (1 x+y)

asuuming the the Lagrangian multipliers for the other constraints are all zero. The partial

derivatives are:

@L@x

= 2 (x 1) = 0@L@y

= 2 (y+ 3) += 0

Eliminating yields (x 1) = y+ 3, y =x 2. Since also y = x 1 is supposedto hold we getx 2 =x 1, 2x= 1 orx= 1

2. This impliesy =1

2. The solution to

the rst order-conditions an complementarity conditions yield (x; y) =12 ;12

, which

satises all constraints. We have= 2 (y + 3) = 2 2:5> 0. Given thatL is concaveinx and y it follows that (x

; y

) =12 ;

12

is the constrained optimum of our problem.

Problem 11 (15 marks) Solve the problem

maxu(t)

Z 10

(x 5)2 dt

subject to _x=u,x (0) = 0,x (1) = 2,1 u 3

Solution 11 The Hamiltonian is

H=

(x

5)2 +pu

To maximizeHwe must have

u (t) =

5ifp (t)> 01 ifp (t) 1. Hence p (t) isstrictly decreasing. This implies thatu (t)can only change sign once, from being positiveand have value +5 to being negative and having value5.

Given the boundary conditions, this only leaves one candidate for x (t), namely

x (t) =

5t ift < t03 t ift t0

where t0 is chosen such that x (t) is continuous at t0, so 5t = 3 t; which means thatt0 =

12

.

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We have hence p12

= 0. Fort < t0

p (t) =

Z 12

t

5 d=5

8 5

2t2

and fort > t0

p (t) =Z t

1

2

(3 ) d= 12

t2 + 3t 118

given that the objective function is concave, this is indeed the optimum.

Part C

Problem 12 (20 marks) Derive the demand function of a consumer with utility function

u (x; y) =p

x +p

y

Solution 12 The utility function is monotonic and so the budget constraint must bebinding. The Lagrangian is

L = px+py+1x+2y+3(bpxxpyy)@L@x

= 1

2p

x+1 3px= 0

@L@y

= 1

2p

y+2 3py = 0

Suppose rst that only the budget constraint binds and so 1 = 2= 0. In this case

1

2p

x = 3px

1

2py = 3py

)p

ypx

= pxpy

py =

pxpy

px

y = p2xp2y

x

Next we use the budget equation.

pxx+px

p2x

p2y x = b

px

p2x+p

2y

p2y

x = b

x =p2y

p2x+p2y

b

px>0

y = p2xp2x+p

2y

b

py>0

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This solution is admissible for any px; py; b > 0. The Lagrangian multiplier 3 for thebudget constraint is positive

3 = 1

2pxp

x >0:

We do not have to consider any further cases, which is of course a bit too simple for a

proper exam question in part C! Because the utility function is easily seen to be concaveand all constraints linear, we have found the optimum for all px; py; b >0. Thus we havedetermined the demand function for both commodities. The case where Lagrangian

e must haveb px py b

Next, if the non-negativity constraint x = 0 is binding we get the optimum x = 0 andy =b=py. For the Lagrange multipliers we obtain 2 = 0,

1

b=py+ 1 =

3py

3 = 1

b +py

1 = 3px 1 = pxb+py

1 0() px py b

This case arises when px py b.Finally, if the non-negativity constraint y = 0 is binding, we must have y = 0 and

x =b=px. Since 1= 0we obtain

1b=px+ 1 = 3px

3 = 1

b+px

2 = 3py 1 = pyb+px

1 0() px py b

Overall we obtain the demand function

(x (px; py; b) ; y (px; py; b)) =

8>:

(p=bx; 0) for px py b

bpx+py2px ; bpy+px2py for b px py b(0;p=by) for b px pyProblem 13 (20 marks) Solve the production planning problem

min

Z T0

c1u

2 +c2x

dt

subject to _x= u,x (0) = 0,x (T) =B; u (t) 0in the caseB < c2T2=4c1, taking explicitaccount of the constraint u 0:Solution 13 see Kamine Schwartz, p. 172f.

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