(1) Given the following system of linear equations, which depends on a parameter a R, 3x y + 5z = 2 4x + y + (a 2 14)z = a + 2

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1 (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 + (a 2 14z = a + 2 (a Classify the system of equations depending on the values of the parameter a. (b Solve the system for the values of the parameter a for which there are infinitely many solutions. (a The matrices associated to the system are a 2 14 a + 2 f 3 f 3 f a 2 16 a 4 f 2 f 2 3 f 1 f 3 f 3 4 f 1 f f a 2 2 a a 2 16 a 4 We see that if a 4 and a 4 then rank A = rank(a B = 3 and the system has a unique solution. If a = 4, then rank A = rank(a B = 2 and the system has infinitely many solutions, described by a parameter. Finally, if a = 4, then rank A = 2 < rank(a B = 3 and the system is not consistent. (b The system of equations has infinitely many solutions for a = 4. For this value, the original original system is equivalent to the following one { x + 2y 3z = 4 whose solution is x = 8 7 y 2z = z, y = 7 + 2z, z R.

2 (2 Consider the set A = {(x, y R 2 : y x + 1, y x + 1, y > } and the function f(x, y = y x 2. (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 (.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. (a The graphic representation of the set A is the following The boundary and the interior of the set A are represented in the following figure The set A is neither open (since, A does not coincide with its interior nor closed (since, A does not contain its boundary. It is bounded, because it is contained in a ball of center (, and radius 2. The set A is not compact, because it is not closed. The set A is convex because A = {(x, y R 2 : g 1 (x, y 1, g 2 (x, y 1, g 3 (x, y > } and the functions g 1 (x, y = x y, g 2 (x, y = y x, g 3 (x, y = y are 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 R 2 : y = x 2 + c} with c R. Graphically, (the arrows point in the direction of growth We see that the global maximum is attained at the point P = (, 1. The maximum value is f(, 1 = 1 = 1. The function f does not attain a global minimum on A.

3 (3 Consider a firm that needs to produce 42 units 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, 1 and 2 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. (a The problem is min s.a. K + 2L K + L = 42 The Lagrange function is L(K, L = K + 2L + λ(42 K L. The Lagrange equations are = L K = 1 = L L = 2 λ 2 K λ 2 L = 42 K L (b The solution is λ = 8, K = 16, L = 4 The cost is K + 2L = 168 monetary units. In order to determine the second order conditions, we calculate the Hessian matrix of L(x, y. 2 L K K = λ 2 L 4 K 3/2, K L = 2 L L K =, 2 L L L = λ 4 L 3/2 Therefore, the Hessian matrix of L at the point λ = 8, K = 16, L = 4 is ( 1 64 H = which is positive definite. Therefore, the point is a local (and global minimum. The savings would be, approximately, λ = 8.

4 (4 Given the following system of equations x 2 y + y 2 z + z 2 x = 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 and z as functions of x in a neighborhood of the point (x, y, z = (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 x = 1. (a Let f 1 (x, y, z = x 2 y + y 2 z + z 2 x 5 and f 2 (x, y, z = x + y + z 2 be the functions that define the restrictions. We have that f 1 and f 2 are polynomials and, hence, C 1 (R 3, We see that the point (1, 1, 2 is a solution of the system of equations. Finally, f 1 f 1 y 2 + 2zx y f 2 y z f 2 z } = x2 + 2yz 1 1 which computed at the point (1, 1, 2 yields = 8 (b The obtain the required derivatives, we differentiate directly with respect to x in the system of equations. We obtain 2xy + x 2 y (x + 2yy (xz + y 2 z (x + 2zz (xx + z 2 = = } 1 + y (x + z (x = which at the point (1, 1, 2 yields } 2 + y (x 4y (x + z (x + 4z (x + 4 = = 1 + y (x + z (x = and solving the linear system of equations we get that z (1 = 5 8, y (1 = 3 8 Taylor s first order approximation of the function y(x in a neighbourhood of the point x = 1 is P 1 (x = y(1 + y (1(x 1 = 1 3 (x 1 8 and Taylor s first order approximation of the function z(x at the same point is Q 1 (x = z(1 + z (1(x 1 = 2 5 (x 1 8

5 (5 Consider the function f(x, y = xye x+2y. (a Determine the critical points (if any of the function f in the set R 2. (b Classify the critical points of f into (local or global maxima, minima and saddle points (a We compute the critical poinbts: f x (x, y = y(1 + xex+2y = f y (x, y = x(1 + 2yex+2y = Solving the above system we obtain the points P (, and Q( 1, 1 2 (b The classify the above points, we compute the Hessian matrix: ( y(2 + xe Hf(x, y = x+2y (1 + x(1 + 2ye x+2y (1 + x(1 + 2ye x+2y x(4 + 4ye x+2y So, ( 1 Hf(, = 1 Therefore, (, is a saddle point, because Hf(, is indefinite (its determinant is 1. Hf ( 1, 1 2 = e 2 ( In this case, the matrix Hf ( 1, 1 2 = is negative definite. Therefore, the point ( 1, 1 2 is a local maximum. It is not a global maximum because lim x f(x, 1 = +.

6 (6 Consider the function f(x, y = 2ax 2 by 2 + 4x 3. Discuss, depending on the values of the parameters a and b, when the function f is strictly convex and/or strictly concave. The gradient vector of f is f(x, y = (4ax + 4, 2by The Hessian matrix of f is Hf(x, y = ( 4a 2b The function would be strictly convex if D 1 > and D 2 > ; and strictly concave if D 1 < and D 2 >, with D 1 = 4a and D 2 = 4a 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 a and b, If 4a > and 2b >, the function f(x, y is strictly convex. If 4a < and 2b <, the function f(x, y is strictly concave. Summarizing, f(x, y is strictly convex in R 2 if a > and b <, and f(x, y is strictly concave in (R 2 if a < and b >.