ECON 11: Problem Set 1 Solutions1. (Unconstrained Optimization-Single Variable) Consider the following function:f (x) = 3x3−5x2+ xwherex ∈ [0,2](a) Find the x’s that are critical value points of f (x) (values of x that are a potential minimum or max-imum). Are those critical values maxima or minima? (Hint: Check the second order derivative.Check also the value of the function at the endpoints 0 and 2 to make sure you have the correctanswer.)Solution:Take FOC to get critical pointsf0(x) = 9x2−10x + 1 = 0 ⇒ x = 1,19Check SOCf00(x) = 18x −10 ⇒ f00(19) < 0, f00(1) > 0Thus, there are two critical points: x =19is a local maximum, x = 1 is a local minimum. But we alsoneed to check for the END POINTSf (0) = 0, f (2) = 6Compare withf (1) = −1, f (19) = −13243Therefore, f (1) = −1, f (2) = 6 are the (global) minimum and maximum respectively.2. (Unconstrained Optimization-Two Variables) Consider the function:f (x1,x2) = x31+ 3x32−9x1x21(a) Find a minimum given that x1,x2≥ 1.Solution: Take FOC: f1= 3x21−9x2= 0, f2= 9x22−9x1= 0We get x1= x2= 0 and x1=3√9,x2=3√3But because x1,x2≥ 1, we only have to consider x1=3√9,x2=3√3Check SOCf11= 6x1= 63√9 > 0 f12= −9f12= −9 f22= 18x2= 183√3 > 0f11f22− f212= 63√9 ×183√3 −(−9) ×(−9) = 182−92> 0Therefore x1=3√9,x2=3√3 is a minimum.Note: the followings is not required and will not be testedFirst, if you are interested, f11f22− f212< 0 when x1= x2= 0. So (0,0) is a saddle point. But youare not required to know what a saddle point is.Also, this function does not have a global maximum or global minimum. For example, x1→∞,x2=0 then f (x1,x2) → ∞; x1→ −∞, x2= 0 then f (x1,x2) → −∞.3. (Constrained Optimization) Julia maximizes the following utility function:u(x1,x2) = x2/71x5/72subject to the budget constraintp1x1+ p2x2= Iwherep1, p2,x1,x2,I > 0(a) Find the (x1,x2) that maximizes u(x1,x2).i. Step 1: Identify the type of problemThis is a CONSTRAINED optimization problem with two variables. Therefore, we need toapply the Lagrangian multiplier method.ii. Step 2: Setup the LagrangianL(x1,x2,λ ) = x2/71x5/72+ λ (I − p1x1− p2x2)iii. Step 3: Derive the FOCs and solve them for the critical pointsL1=27x−5/71x5/72−λ p1= 0 ⇒27x−5/71x5/72= λ p1(1)L2=57x2/71x−2/72−λ p2= 0 ⇒57x2/71x−2/72= λ p2(2)L3= I − p1x1− p2x2= 0 ⇒ p1x1+ p2x2= I (3)From the above two equations we get(1)(2)⇒2x25x1=p1p2⇒ x2=5x1p12p22substituting into equation (3), we getx∗1=2I7p1x∗2=5I7p2iv. The SOCs for a constrained optimization problem are not required and will not be tested, so Iwill omit them here, but the critical point is a maximizer.(b) Show that the maximizer x∗2(p1, p2,I) is decreasing in p2and increasing in I (Hint: Use the partialderivatives).This question asks to compute the change in x∗2as p2and I increase. Since the change in the func-tion x∗2with respect to one of the variables is determined by the partial derivative, this question isessentially asking you to compute the derivatives and to check their sign.∂ x∗2∂ p2= −5I7p22< 0∂ x∗2∂ I=57p2> 0(c) Does x∗2(p1, p2,I) change if p1changes? (Hint: partial derivatives).∂ x∗2∂ p1= 0Therefore, x∗2(p1, p2,I) does not change if