(1) Given the following system of linear equations, which depends on a parameter a∈R,
x+ 2y−3z = 4
3x−y+ 5z = 2
4x+y+ (a2−14)z = a+ 2
(a) Classify the system of equations depending on the values of the parametera.
(b) Solve the system for the values of the parameter afor which there are infinitely many solutions.
Soluci´on:
(a) The matrices associated to the system are
1 2 −3 4
3 −1 5 2
4 1 a2−14 a+ 2
f27→f2−3×f1
f37→f3−4×f1
7→
1 2 −3 4
0 −7 14 −10
0 −7 a2−2 a−14
f37→f3−f2
7→
1 2 −3 4
0 −7 14 −10
0 0 a2−16 a−4
f27→−17×f2
7→
1 2 −3 4
0 1 −2 107
0 0 a2−16 a−4
We see that if a 6= 4 and a 6= −4 then rankA = rank(A|B) = 3 and the system has a unique solution.
Ifa= 4, then rankA= rank(A|B) = 2 and the system has infinitely many solutions, described by a parameter.
Finally, ifa=−4, then rankA= 2<rank(A|B) = 3 and the system is not consistent.
(b) The system of equations has infinitely many solutions fora= 4. For this value, the original original system is equivalent to the following one
x+ 2y−3z = 4 y−2z = 107 whose solution isx= 87−z,y=107 + 2z,z∈R.
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(2) Consider the set A={(x, y)∈R2:y≤x+ 1, y≤ −x+ 1, y >0} and the functionf(x, y) =y−x2. (a) Draw the set A, its boundary and its interior. Determine, justifying the answers, if the set A is
closed, open, bounded, compact and/or convex.
(b) (0.5 points) Do the hypotheses of Weierstrass’ Theorem hold? Why?
(c) (1.5 points) Draw the level curves of f. Indicate the direction of growth. Using its level curves, determine if the function f attains a global maximum (and/or a minimum) value in the set A.
Find the points at which it is attained.
Soluci´on:
(a) The graphic representation of the set Ais the following
(-1,0) (1,0)
(0,1)
The boundary and the interior of the setAare represented in the following figure
(-1,0) (1,0)
(0,1)
(-1,0) (1,0)
(0,1)
The setAis neither open (since,Adoes not coincide with its interior) nor closed (since,Adoes not contain its boundary). It is bounded, because it is contained in a ball of center (0,0) and radius 2.
The setAis not compact, because it is not closed. The set Ais convex because A={(x, y)∈R2:g1(x, y)≥ −1, g2(x, y)≥ −1, g3(x, y)>0}
and the functionsg1(x, y) =x−y,g2(x, y) =−y−x,g3(x, y) =yare linear and, therefore, convex.
(b) The assumptions of Weierstrass’ Theorem do not hold because the set A is not compact (it is bounded, but not closed).
(c) The level curves of f are the sets {(x, y)∈R2:y=x2+c} withc∈R. Graphically, (the arrows point in the direction of growth)
(-1,0) (1,0)
(0,1)
We see that the global maximum is attained at the point P = (0,1). The maximum value is f(0,1) = 1−0 = 1. The functionf does not attain a global minimum onA.
(3) Consider a firm that needs to produce 42units of a certain product at the minimum possible cost. If the firm uses K units of capital and L units of labour, its production is √
K+√
L units. The prices per unit of capital and labour are, respectively, 1and20 monetary units.
(a) Write the optimization problem of the firm. Write the associated Lagrangian funtion and the Lagrange equations.
(b) Solve the Lagrange equations. Check the second order conditions for the critical points and find the solution of the problem.
Suppose now that the firm needs to produce 41 units. Using the computations you have already done and without solving the problem again, determine approximately how many monetary units would the company save in this case.
Soluci´on:
(a) The problem is
min K+ 20L
s.a. √ K+√
L= 42 The Lagrange function isL(K, L) =K+ 20L+λ(42−√
K−√
L). The Lagrange equations are
0 = ∂L
∂K = 1− λ 2√ K
0 = ∂L
∂L= 20− λ 2√ L 0 = 42−√
K−√ L (b) The solution is
λ= 80, K= 1600, L= 4
The cost isK+ 20L= 1680 monetary units. In order to determine the second order conditions, we calculate the Hessian matrix ofL(x, y).
∂2L
∂K∂K = λ
4K−3/2, ∂2L
∂K∂L = ∂2L
∂L∂K = 0, ∂2L
∂L∂L = λ 4L−3/2 Therefore, the Hessian matrix ofLat the pointλ= 80, K= 1600, L= 4 is
H = 20 1
64000 0
0 1
16√ 2
which is positive definite. Therefore, the point is a local (and global) minimum.
The savings would be, approximately,λ= 80.
(4) Given the following system of equations
x2y+y2z+z2x = 5 x+y+z = 2
(a) Apply the Implicit Function Theorem (checking that the hypotheses are satisfied) to show that, in the above system of equations, it is possible to obtain the variables y andz as functions ofxin a neighborhood of the point (x0, y0, z0) = (1,−1,2), so that x and the functions y(x) and z(x) are solutions of the above system of equations that satisfy y(1) =−1,z(1) = 2.
(b) Compute Taylor’s first order approximation of the functions y(x),z(x)around the point x0= 1.
Soluci´on:
(a) Let f1(x, y, z) = x2y+y2z+z2x−5 and f2(x, y, z) =x+y+z−2 be the functions that define the restrictions. We have that
• f1andf2are polynomials and, hence,C1(R3),
• We see that the point (1,−1,2) is a solution of the system of equations.
• Finally,
∂f1
∂y
∂f1
∂z
∂f2
∂y
∂f2
∂z
=
x2+ 2yz y2+ 2zx
1 1
which computed at the point (1,−1,2) yields
−3 5
1 1
=−86= 0
(b) The obtain the required derivatives, we differentiate directly with respect tox in the system of equations. We obtain
2xy+x2y0(x) + 2yy0(x)z+y2z0(x) + 2zz0(x)x+z2= 0 = 0 1 +y0(x) +z0(x) = 0
which at the point (1,−1,2) yields
−2 +y0(x)−4y0(x) +z0(x) + 4z0(x) + 4 = 0 = 0 1 +y0(x) +z0(x) = 0
and solving the linear system of equations we get that z0(1) =−5
8, y0(1) =−3 8
Taylor’s first order approximation of the functiony(x) in a neighbourhood of the pointx0= 1 is P1(x) =y(1) +y0(1)(x−1) =−1−3
8(x−1)
and Taylor’s first order approximation of the functionz(x) at the same point is Q1(x) =z(1) +z0(1)(x−1) = 2−5
8(x−1)
(5) Consider the function f(x, y) =xyex+2y.
(a) Determine the critical points (if any) of the functionf in the setR2.
(b) Classify the critical points of f into (local or global) maxima, minima and saddle points Soluci´on:
(a) We compute the critical poinbts:
∂f
∂x(x, y) =y(1 +x)ex+2y= 0
∂f
∂y(x, y) =x(1 + 2y)ex+2y= 0
Solving the above system we obtain the pointsP(0,0) andQ(−1,−12) (b) The classify the above points, we compute the Hessian matrix:
Hf(x, y) =
y(2 +x)ex+2y (1 +x)(1 + 2y)ex+2y (1 +x)(1 + 2y)ex+2y x(4 + 4y)ex+2y
So,
Hf(0,0) =
0 1 1 0
Therefore, (0,0) is a saddle point, becauseHf(0,0) is indefinite (its determinant is−1).
Hf
−1,−1 2
=e−2
−12 0
0 −2
In this case, the matrix Hf −1,−12
= is negative definite. Therefore, the point −1,−12 is a local maximum. It is not a global maximum because limx→∞f(x,1) = +∞.
(6) Consider the functionf(x, y) = 2ax2−by2+ 4x−3. Discuss, depending on the values of the parameters a andb, when the function f is strictly convex and/or strictly concave.
Soluci´on: The gradient vector off is
5f(x, y) = (4ax+ 4,−2by) The Hessian matrix off is
Hf(x, y) =
4a 0 0 −2b
The function would be strictly convex ifD1>0 andD2>0; and strictly concave ifD1<0 andD2>0, withD1= 4aand
D2=
4a 0
0 −2b
Given that the Hessian matrix is diagonal, it is strictly convex if diagonal terms are positive and strictly concave it they are negative.
In terms of the parameters aandb,
• If 4a>0 and−2b >0, the functionf(x, y) is strictly convex.
• If 4a<0 and−2b <0, the functionf(x, y) is strictly concave.
Summarizing, f(x, y) is strictly convex in R2 if a >0 and b < 0, andf(x, y) is strictly concave in (R2 ifa <0andb >0.
