Calculation and Analysis of Functions and Contour Lines, Ejercicios de Matemáticas

¡Descarga Calculation and Analysis of Functions and Contour Lines y más Ejercicios en PDF de Matemáticas solo en Docsity! Mathematics II Bachelor in Economics Bachelor in Business Administration Problem Set 2: Functions of several variables Solutions. Departament d’Economia i d’Història Econòmica 1 1. Consider the functions: (a) f(x, y) = x2 − sin xy; (b) g(x, y, z) = (x− y, y + ln z) Determine f(1, 0), f(a, b); g(0,−1, e), g(t, t+ 1, t− 1) Solution (a) f(1, 0) = 12 − sin(0 · 1) = 1− sin(0) = 1 f(a, b) = a2 − sin(ab) (b) g(0,−1, e) = (0− (−1),−1 + ln e) = (1, 0) g(t, t+ 1, t− 1) = (t− (t+ 1), (t+ 1) + ln(t− 1)) = (−1, (t+ 1) + ln(t− 1)) 2. Find the (natural) domain of the following functions and describe these sets: (a) f(x, y) = x2 − lnxy; (b) f(x, y) = √ x2 + y2 − 1; (c) f(x, y) = √ x+ √ y (d) f(x, y) = √ x− y; (e) f(x, y) = √ xsin y; (f) f(x, y) = ex/y (g) f(x, y) = ln x y ; (h) f(x, y) = xe1/y Solution (a) Open set, unbounded, non-compact, non-convex. Df = {(x, y) ∈ IR2|xy > 0} = {(x, y) ∈ IR2|x > 0, y > 0} ∪ {(x, y) ∈ IR2|x < 0, y < 0} 2 (e) Open set, unbounded, non-compact, non-convex. Df = {(x, y) ∈ IR2|y 6= 0} (f) Open set, unbounded, non-convex. Df = {(x, y) ∈ IR2|x y > 0, y 6= 0} 5 (g) Is the same graph as in (f). Df = {(x, y) ∈ IR2|y 6= 0} 3. Determine analytically (using symbols) and geometrically (using a graph) the contour lines of the following functions from IR2 into IR (a) f(x, y) = x+ y; (b) f(x, y) = x− 2y; (c) f(x, y) = x2 − 2y (d) f(x, y) = x− y2; (e) f(x, y) = x2 + y2; (f) f(x, y) = x2 − y2 (g) f(x, y) = min{x, y}; (h) f(x, y) = ln(x2 + y2); (i) f(x, y) = x2y Solution To represent an arbitrary function z = f(x, y), cut its graph with horizontal planes z = k, that is parallel to the plane xy. Project these intersections perpendicularly onto the plane xy drawing the contour lines. Therefore, the contour lines are equations f(x, y) = k. (a) f(x, y) = x+ y ⇒ k = x+ y ⇒ y = −x+ k, are lines parallel to the −45-degree line. See this link for Geogebra code. 6 (b) f(x, y) = x − 2y ⇒ k = x − 2y ⇒ y = x− k 2 , are lines parallel to the line y = x/2. (c) f(x, y) = x2−y ⇒ k = x2−2y ⇒ y = x2 − k 2 , are parabolic curves. 7 (i) f(x, y) = x2y ⇒ k = x2y ⇒ y = k x2 . They are hyperbolic curves. 4. Draw the contour lines for the following functions, and the given set. Does there exist a maximum and/or minimum of the function on the set? If so, find them. (a) f(x, y) = 3x− y on A = {(x, y) ∈ IR2|x ≥ 0, y ≥ 0, y + x− 5 ≤ 0} (b) f(x, y) = 2x− y on A = {(x, y) ∈ IR2|x2 + y2 = 4} (c) f(x, y) = x+ y on A = {(x, y) ∈ IR2|y − 3x− 9 ≥ 0} (d) f(x, y) = min{x, y} on A = {(x, y) ∈ IR2|x ≥ 0, y ≥ 0, y + 5x− 10 ≤ 0} Solution (a) Using Weierstrass’ theorem we can justify the existence of maxima and minima because the set A is compact and the function f is contin- uous (it is a polynomial) throughout its domain IR2. The contour lines are: k = 3x − y ⇒ y = 3x − k. These are lines parallel to the line y = 3x. • The absolute maximum is located at (5, 0). Substituting this point in the function f(x = 5, y = 0) = 3 · 5 − 0 = 15 = k we find k = 15. Therefore, kmax = 15. • The absolute minimum is located at (0, 5). Substituting this point in the function f(x = 0, y = 5) = 3 · 0 − 5 = −5 = k we find k = −5. Therefore, kmin = −5. • Geogebra code to explore this example is at https://www.geogebra.org/graphing/tyjwz4sq 10 (b) Using Weierstrass’ theorem we can justify the existence of maxima and minima because the set A is compact and the function f is contin- uous (it is a polynomial) throughout its domain IR2. The contour lines are: k = 2x − y ⇒ y = 2x − k. These are lines parallel to the line y = 2x. We can identify these maximum and minimum points. Looking at the figure we see that these points lay on the line orthogonal to the line y = 2x. Thus, we identify the intersection between the circle and the 11 line y = −(1/2)x,y = −1 2 x x2 + y2 = 4 ⇒ x2 + (−1 2 x )2 = 4⇔ 5x2 = 16⇔ x = ± √ 16 5 to obtain: The absolute maximum on A is ( x = 4√ 5 , y = −2√ 5 ) with value f ( x = 4√ 5 , y = −2√ 5 ) = 2 4√ 5 − −2√ 5 = k = 2 √ 5 ≈ 4.47 The absolute minimum on A is ( x = −4√ 5 , y = 2√ 5 ) with value f ( x = −4√ 5 , y = 2√ 5 ) = 2−4√ 5 − 2√ 5 = k = −2 √ 5 ≈ −4.47 Alternatively, we could use the fact that the extreme contour lines are tangent to the circle. This means that at the tangency point the slopes of the two functions are equal. To compute the slope of the circle x2 + y2 = 4 let us first rewrite it as y = (4− x2) 1 2 . Then, dy dx = 1 2 (4− x2)− 1 2 (−2x) = −x√ (4− x2 The slope of the lines y = 2x − k is 2. Therefore, we are looking for the value of x solution of −x√ (4− x2 = 2. So, −x = 2 √ 4− x2 Square both sides to get x2 = 4(4− x2) = 16− 4x2 Set z = x2, so that z = 16− 4z which gives z = 16 5 . Therefore, x = ± 4√ 5 , as we obtained before. 12 The figure shows the contour lines and the set A that is a disk with center at (5, 0) and radius 1. The tangency points between the contour lines and the set A lay in the x-axis at points (4, 0) and (6, 0). • Absolute minimum on A is (6, 0) with value kMIN = f(6, 0) = 1 36 . • Absolute maximum on A is (4, 0) with value kMAX = f(4, 0) = 1 16 . 6. Use level curves to geometrically find the absolute minimum and maximum of the following functions in the proposed domains. (a) f(x, y) = xy in the domain x+ y ≤ 1, x, y > 0 (b) f(x, y) = x− y2 in the domain x2 + y2 ≤ 2 (c) f(x, y) = min{x, y} in the domain x+ 2y ≤ 1, x, y > 0 Solution Theorem 1 (Weierstrass). Let f : A ⊂ IRn → IR be continuous on A. Let A ⊂ D be compact. Then, f is bounded in A and ∃ x1, x2 ∈ A such that x1 is the global maximum of f in A, and x2 is the global minimum of f in A. (a) The figure shows the contour lines y = k/x and the set A. We cannot apply Weierstrass’ theorem because A is not closed and thus it is not compact. If x, y ≤ 0 then A would be bounded and the minimum would be f(0, 0) = 0 = f(x, 0) = f(0, y); the maximum would be found on the line x+ y = 1, at the point (0.5, 0.5). (b) In this case the contour lines have the form y = √ x− k. The set A is bounded and the function f is continuous on A. Thus, Weierstrass theorem ensures that there is a global minimum and a global maximum of f on A: • Absolute maximum on A is f( √ 2, 0) = √ 2. 15 • Absolute minimum on A is f(−1 2 ,± √ 7 4 ) = −9 4 = −2.25. (c) We cannot apply Weierstrass’ theorem because A is not closed and thus it is not compact. However, we can identify the global maximum and the global minimum combining the following equations:{ x = y x+ 2y = 1 → { ymax = 1 3 xmax = 1 3 16 7. Mathematics may help to think about economic problems systematically. Try to express the following economic problems in mathematical terms. (a) A bakery specializes in three different types of bread: ciabatta, olive baguette and wholegrain. Each ciabatta sells for 1C, olive baguettes for 2C and each wholegrain bread for 1.5C. The owner wants to de- termine the optimal amount of production to get the maximum daily benefit. However, there are budget and time constraints. The cost of a piece of ciabatta is 0.25C, a piece of olive baguette is 0.75C and a wholegrain bread is 0.50C. Daily expenditures must be lower than 300C. However, the preparation and cooking time varies with the type of bread. To make 5 bars of ciabatta it takes 8 minutes to, to make 3 olive baguettes 10 minutes and to make 2 wholegrain breads it takes 9 minutes. And the total time that can be used to prepare all products must not exceed 4 hours. (b) A company has decided to reduce the costs of their employees because it is having a bad time. Instead of reducing the privileges of their employees such as the coffee machines or free tickets for lunch, it has decided to reduce the workforce. There are three types of workers who are classified with respect to their level of education: Masters, Bachelors, and Interns. There are 100 Bachelors, 150 Masters and 50 Interns. Costs for each type are respectively 10C, 20C and 15C. The revenues are: 20C for Bachelors, 25C for Master and 15C for Interns. The company needs to take measures to make profits of at least 8000C. This measure should not affect more than 100 people. How many employees of each type should the firm fire? Solution (a) The following table summarizes the data of the problem: Production (1 unit) Price Cost Time Profit Ciabatta (x) 1 0.25 5x→ 8 min 0.75 Olive (y) 2 0.75 3y→ 10 min 1.25 Wholegrain (z) 1.5 0.5 2z→ 9 min 1 Also, there are budget and time constraints. Let x denote the daily quantity of ciabatta bread; let y be the daily quantity of olive baguettes; and let z denote the daily production of wholegrain bread. Then, 17 or y = (U0 x 1 2 )2 = U2 0 x The slope of this representative level set is dy dx = −U 2 0 x2 < 0,∀x > 0 Also, d2y dx2 = 2U0 x3 > 0 Therefore, the contour lines are decreasing and convex. Also, they are asymptotic to infinite both in y and x (verifying this is left to the reader). The budget set is defined as the bundles of goods (x, y) affordable to the individual. This is, B = {(x, y)|pxx+ pyy ≤ w} We are interested in the frontier of this set, the so called budget con- straint, namely pxx+ pyy = w in our example this reduces to x−2y = 50, or y = 25− 1 2 x. The slope of this budget constraint is −1 2 . Thus, this is a straight line. The following figure represents all this information. x y 50 25 0 x∗ y∗ 20 (b) The consumption bundle maximizing the individual’s utility is char- acterized by the tangency between a contour line and the budget con- straint. This is, −U 2 0 x2 = −1 2 or x = U0 √ 2 Substituting the value of x∗ in the budget constraint we obtain y = 25− U0√ 2 Substituting these values of x and y in the utility function we obtain the value of U0: U0 = (U0 √ 2) 1 2 (25− U0√ 2 ) 1 2 yielding U0 = 25√ 2 Finally, substituting the value of U0 in x and y we identify the optimal (utility-maximizing) bundle: x∗ = 25; y∗ = 25 2 9. Consider a firm that using one input (labor, l) to produce a consumption good y. The technology is described by a production function y = f(l). Suppose f is continuous and strictly concave. Let w be the price of one hour of labor (wage). Let p be the prices of the output. Both p and w are exogenous. The problem of the firm is to determine the hours of work to maximize profits. (a) Represent graphically the problem of the firm. (b) Compute analytically the optimal hours of work to hire. Solution (a) To identify the optimal number of hours we need to plot the isoprofit map and the production function as in the following figure: 21 y y∗ l∗ l (b) The profit function of the firm is π(l) = py −wl, so that a representa- tive isoprofit curve is y = π0 p + w p l Therefore, the isoprofit map is a set of increasing straight lines with slope w/p. Note that the intercept of these lines with the y-axis is the level of profits π0/p. This means that higher isoprofit lines are associ- ated with higher profit levels. The highest feasible isoprofit line is the one tangent to the production function. At that point the slopes of of the two curves coincide. Accordingly the optimal (profit maximizing) level of hours to contract is given by the solution of dy dl = w p . 10. A firm uses two inputs (k.l) to produce a consumption good. Let y = f(k, l) denote the production technology. In accordance with the instructions re- ceived in the general assembly of shareholders, the objective of the firm is to produce y units of the good. The problem of the firm is to achieve that objective with the minimum cost. Let (r, w) denote the prices of the in- puts (k.l) respectively. Suppose that firm faces a fixed cost c0. Suppose the technology f generates strictly convex isoquants. (a) Represent graphically the problem of the firm. (b) Compute analytically the optimal demand of inputs. Solution 22