Final Exam - Answer key

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1 Fall4 Final Exam - Answer key ARE Problem (Analysis) [7 points]: A) [ points] Let A = (,) R be endowed with the Euclidean metric. Consider the set O = {( n,) : n N}. Does the open cover O of A have a finite open subcover? If so, provide it. If not, prove that one does not exist. Ans: This open cover does not have a finite open subcover. Want to show that if we take any finite subset of O we will no longer be able to cover A. Let F be an aribitrary, finite set of natural numbers and consider O F = {( n,) : n F}. Since F is finite there must be a largest element of the set. Denote this as ˆn. Hence O F = ( ˆn,). Now consider the point a = ˆn. Clearly a A, but a / O F. Hence O F does not cover A.

2 a): det(a) a): det([a(:,)),b]..4.6 a): det([b,a(:,))] b): det(a) b): det([a(:,)),b] b): det([b,a(:,))] c): det(a) c): det([a(:,)),b] c): det([b,a(:,))] Problem (Linear Algebra) [5 points]: Figure. Figure for problem A) [ points] Suppose that X is a vector space and that A is a subset of X. Prove that if A is a vector subspace then X/A is not a vector subspace. Ans: All vector spaces (and subspaces) must be closed under vector addition and scalar multiplication. A consequence of this is that all vector spaces must contain the zero vector. If A is a vector subspace then we must have A, hence / X\A. Therefore X\A cannot be a vector subspace of X. B) [5 points] For each of panels a), b) and c) of figure, write down your best guess for the solution of the equation Ax = b, where A = [v v ]. (Note: full marks will be awarded if the maximum element-wise distance between your answer and the true answer is <.. Otherwise the further your answer is from the correct answer, the fewer marks you will recieve. For full marks use only the graphical approach to Cramer s rule (no algebra). Hint: be careful about the sign of your solution.) Ans: See 4th Linear Algebra Notes for full discussion of the graphical interpretation of Cramer s rule. a) x =.8; x =. b) x = ; x = c) x = ; x =

3 Problem 3 (Calculus) [5 points]: Let f(x) = x 3 +x 3x. A) [9 points] Use a first order Taylor expansion around x = to estimate the change in the function when x is increased by, i.e. dx =. What is the estimated sign of the change? Ans: Note that for f(x) = x 3 +x 3x, f (x) = 3x +4x 3, f (x) = 6x+4, and f (x) = 6. The first order approximation of the change in the function is given by f(x + dx) f(x ) = f (x )dx = 3dx. For dx = the estimated change is 6 and the sign of the estimated change is negative. B) [9 points] Now use a second order Taylor expansion around x = to estimate the change in the function when x is increased by, i.e. dx =. What is the estimated sign of the change? Ans: The second order approximation of the change in the function is given by f(x + dx) f(x ) = f (x )dx+ f (x ) dx = 3dx+dx. For dx = the estimated change is and the sign of the estimated change is positive. C) [7 points] Evaluate the original function at x =. Discuss how the actual sign of the change compares to your answers in the previous two parts. Ans: The orignal function evaluated at x = is f() =, hence the actual change in function, f() f() =, which is positive. This is an example of the importance of the condition in the Taylor-Young theorem that the sign of the estimated change is equal to the sign of the actual change for a sufficiently small dx. Here the dx is too large for the first order approximation to yield the correct sign, but the dx is sufficiently small for the second order approximation to give the correct result. 3

4 4 s s s s s s Figure. Figure for problem 4b Problem 4 (Constrained Optimization) [5 points]: After working hard all semester you decide to take a vacation by the seaside. Each day you can either spend your time sunbaking, s, or jetboating, j. Consuming one unit of either takes a day. Your vacation is 5 days long. Additionally sunbaking costs $ per day and jetboating costs $4 per day. You have allocated a budget of 4 dollars for your vacation. Your utility from sunbaking and jetboating is given by f(s,j) = sj. A) [3 points] Set up your optimization problem so you have the best time possible on vacation (please ignore non-negativity constraints). Ans: The vacation-fun maximization problem is given by maxsj subject to g : s+4j 4 g : s+j 5 (Strictly speaking this problem should have non-negativity constraints on s and j. We omit them in this problem since it increases the number of cases unnecessarily. We will only consider (s,j) R ++ R ++ to avoid the discontinuity of f at (,).) B) [3 points] Sketch the constraint set and draw a level set of the utility function. Ans: C) [3 points] Write down the Lagrange and the KKT conditions. Ans: The Lagrangian is given by L = sj +λ (4 s 4j)+λ (5 s j)

5 5 The KKT conditions are: () L s = j λ λ = () L j = s 4λ λ = (3) L λ = 4 s 4j (6)λ (4 s 4j) = (9)λ (4) L λ = 5 s j (7)λ (5 s j) = ()λ D) [8 points] Find the solution to the problem. Show all your work for maximum credit. Ans: The constraints are linear so the constraint qualification holds globally. Case : Interior solution, λ =,λ = f = (j,s) The gradient is never vanishing on R ++ R ++. No interior solution. Case : λ >,λ > g s+4j = 4 g s+j = 5 3j = 9 j = 3 There is one candidate point here (,3). () 3 λ λ = () 4λ λ = 9 = 3λ λ = 3 λ = Contradiction. Case 3: λ =,λ > () j = λ () s = λ j = s (4) 5 s j = 5 s = s = j = 5 (3) = 3.5(< ) Contradiction. Case 4: λ >,λ = () j = λ () s = 4λ = 4j (3) 4 4j 4j = j = 3 s = λ = 3 (4) 5 3 = ( ) Solution. (s,j ) = (,3),f(,3) = 36 E) [8 points] What will be the approximate increase in your utility if: (a) Your holiday is extended for a day? (b) You find a dollar on the beach? Ans: The Lagrangian multiplier represents the first order approximation of the increase in the value function if the corresponding constraint is relaxed by unit. (a) TheLagrangemultipliercorrespondingtothetimeconstraint, λ, is(satisfiedwithequality, but not binding). Hence extending your holiday by a day will not increase your utility, nor will it change your choice of s and j.

6 6 (b) The Lagrange multiplier corresponding to the time budget constraint, λ, is 3. Hence, to a first order approximation, your utility will increase by 3 utils if you find a dollar on the beach.

7 7 s t Figure 3. Figure for problem 5 Problem 5 (Comparative Statics) [5 points]: Consider the following constrained maximization problem. maxx+α subject to x 3 x α A) [5 points] In α x space, draw the feasible set and a level set of the objective function. Ans: Note, the way the question was posed in the exam (with the α = condition) did not make sense. Everyone got full points for this question. B) [ points] Use the implicit function theorem to calculate dx dα. Ans: The Lagrangian of this problem is give by L(x,λ;α) = x+α+λ(α x 3 x) The first order conditions are: L x = λ(3x ) = L λ = (α x3 x) = Let [ λ(3x f = ] ) α x 3 x The implicit function theorem gives us that [ x ] [ f λ = Jf(x,λ) f ]

8 8 Jf(x,λ) = [ f x f x f λ f λ ]=[ 6λx 3x ] 3x Notice that det(jf) = (3x +), which is always negative, hence we can apply the implicit function everywhere Using Cramer s rule which is always positive. x = [ 3x det(jf) ] = 3x + C) [ points] Use the envelope theorem to estimate the maximized value of the objective function when α =.9. Ans: The maximized value can be approximated by f(x (α))+ df(x (α)) dα dα when α = and dα =.. x (α = ) =, λ (α = ) = 4 from the first order conditions. From the envelope theorem, df(x (α)) dα = f(x (α)) = +λ (α) = 3 4 The maximized objective function is then f(x (α))+ df(x (α)) dα = +()+ 5 dα 4 +λ (α) g(x (α)) ( ) = = 3 8 =.875