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1 Maths Review Questions - Set 2 1.1 Partial Derivatives - one more try Take all the partial derivatives of the following functions 1. f (x, y, z) = x3 y 2 z + z 2 y + y 2 x fx (x, y, z) = 3x2 y 2 z + y 2 fy (x, y, z) = 2x3 yz + z 2 + 2yx fz (x, y, z) = x2 y 2 + 2zy 2. f (x, y, z) = (xy + z)(x + y)3 fx (x, y, z) = y(x + y)3 + 3(x + y)2 (xy + z) fy (x, y, z) = x(x + y)3 + 3(x + y)2 (xy + z) fz (x, y, z) = (x + y)3 3. f (x, y) = log xy There are two ways to solve this. One way is to note that f (x, y) = log xy = log x + log y =⇒ fx (x, y) = fy (x, y) = 1 x 1 y Or by chain rule 1 fx (x, y) = y. yx = 1 x 1 fy (x, y) = x. yx = 1 y 4. f (x, y) = log xy As above fx (x, y) = 1 x fy (x, y) = − y1 3 5. f (x, y) = (x2 + (xy + y 2 )) 2 1 fx (x, y) = 32 (x2 + (xy + y 2 )) 2 (2x + y) 1 fy (x, y) = 32 (x2 + (xy + y 2 )) 2 (2y + x) 1.2 Maximizing functions of one variable For each of the following functions • Find the critical points (i.e. the points where the function is flat) • Categorize these points as local maxima or minima (if possible) 1 • Explain whether any of these points are global maxima or minima Take the first derivatives of the following functions 1. f (x) = −3x2 + 9x + 8 FOC: f 0 (x) = −6x + 9 = 0 x= 3 2 This is the only critical point SOC: f 00 (x) = −6 < 0 This is therefore a local maximum As we only have one critical point, and it is a local maximum, then we know from theorem 4 from the Prelim handout 2 that we have a global maximum. 2. f (x) = 7x2 + 2x + 3 FOC: f 0 (x) = 14x + 2 = 0 x = − 17 This is the only critical point SOC: f 00 (x) = 14 > 0 This is therefore a local minimum As we only have one critical point, and it is a local minimum, then we know from theorem 4 from the Prelim handout 2 that we have a global minimum. 3. f (x) = x3 3 FOC: + x2 − 8x + 10 f 0 (x) = x2 + 2x − 8 = (x + 4)(x − 2) = 0 x = −4, 2 There are two critical points SOC: f 00 (x) = 2x + 2 f 00 (−4) = −6 < 0 This is a local maximum f 00 (2) = 6 > 0 2 This is therefore a local minimum We can see that neither of these points are global extrema. The easiest way to see this is to note that, as x gets very large, f (x) goes to +∞ and as x gets very small, f (x) goes to −∞ 1.3 Maximizing functions of two variables For the following functions, find the critical points (i.e. the point where the function is flat with regard to all its arguments) 1. f (x, y) = x4 + x2 − 6xy + 3y 2 FOC fy (x, y) = −6x + 6y = 0 ⇒x=y fx (x, y) = 4x3 + 2x − 6y = 0 but as x=y ⇒ 4x3 + 2x − 6x = 0 ⇒ 4x(x2 − 1) = 0 ⇒ x = 0, x = 1, x = −1 So critical points lie at (0, 0), (1, 1), (−1, −1) 2. f (x, y) = xy 2 + x3 y − xy FOC fx (x, y) = y 2 + 3x2 y − y = 0 ⇒ y(y + 3x2 − 1) = 0 fy (x, y) = 2xy + x3 − x = 0 ⇒ x(2y + x2 − 1) = 0 The FOC with respect to x tells us that either y = 0 or y + 3x2 − 1 = 0. Similarly, the second equation tells us that either x = 0 or 2y +x2 −1 = 0 . This gives us 4 different cases (i) x = 0, y = 0 (ii) y = 0, 2y + x2 − 1 = 0 ⇒ x2 = 1 ⇒ x = 1, x = −1 (iii) y + 3x2 − 1 = 0, x = 0 ⇒y=1 (iv) y + 3x2 − 1 = 0, 2y + x2 − 1 = 0 3 ⇒ x2 = 1 − 2y ⇒ y + 3(1 − 2y) − 1 = 0 ⇒ 5y = 2 ⇒ y = 2 5 ⇒ x2 = 1 − 2. 25 = ⇒x= √1 , x 5 = 1 5 − √15 So, collecting all the different cases, we get the following critical points (0, 0), (−1, 1), (1, 1), (0, 1), (− √15 , 25 ), ( √15 , 25 ) 1.4 Constrained optimization In the following two problems there is only one critical point, which I shall tell you is the maximum. Find it. 1. Maximize 3 log x + 2 log y subject to 3x + 4y = 10 (i) Form the Lagrangian L(x, y, µ) = 3 log x + 2 log y − µ(3x + 4y − 10) (ii) Take the first order conditions Lx (x, y, µ) = Ly (x, y, µ) = 3 x 2 y − 3µ = 0 ⇒ − 4µ = 0 ⇒ 3 x 2 y = 3µ = 4µ Lµ (x, y, µ) = 3x + 4y − 10 = 0 (iii) Divide the x, y FOC by each other to get rid of µ 3 x 2 y = 3µ 4µ ⇒ 3y 2x = 3 4 (iv) Get x in terms of y (or visa versa) 3y 2x = 3 4 ⇒y= x 2 (v) Substitute into the constraint and solve 3x + 4( x2 ) = 10 ⇒ 5x = 10 ⇒ x = 2 ⇒ y = 1 2. Maximize xα y (1−α) subject to px x + py y = I.(Note that a, px , py I are parameters for this problem your answer should be a function of x and y in terms of these values) See separate sheet If you use the Lagrangian method on the following problem, you will get two solutions. One will be a maximum and the other will be a minimum. Find them both and decide which is which. First, a quick apology - there are actually 4 solutions to this problem, but as we shall see below, they come in two very similar pairs of two. 4 1. Maximize x2 + y 2 subject to x2 + xy + y 2 = 3 (i) Form the Lagrangian L(x, y, µ) = x2 + y 2 − µ(x2 + xy + y 2 − 3) (ii) Take the first order conditions Lx (x, y, µ) = 2x − µ(2x + y) = 0 ⇒ 2x = µ(2x + y) Ly (x, y, µ) = 2y − µ(2y + x) = 0 ⇒ 2y = µ(2y + x) Lµ (x, y, µ) = x2 + xy + y 2 − 3 = 0 (iii) Divide the x, y FOC by each other to get rid of µ 2x 2y = µ(2x+y) µ(2x+y) ⇒ x y = (2x+y) (2y+x) (iv) Get x in terms of y (or visa versa) x y = (2x+y) (2y+x) ⇒ x(2y + x) = y(2x + y) ⇒ 2xy + x2 = 2xy + y 2 ⇒ x2 = y 2 ⇒ x = y or x = −y (v) Substitute into the constraint and solve Case 1: x = y x2 + xy + y 2 = 3 ⇒ x2 + x2 + x2 = 3 ⇒ 3x2 = 3 ⇒ x2 = 1 Which gives 2 solutions: (1, 1) and (−1, −1) Case 2: x = −y x2 + xy + y 2 = 3 ⇒ x2 − x2 + x2 = 3 ⇒ x2 = 3 √ √ √ √ Which gives 2 solutions: (− 3, 3) and ( 3, − 3) To find out which of these are the max and which are the minima, substitute the solutions back into the objective function. This tells you that the two solutions from case 1 give you f (x, y) = x2 + y 2 = 1 + 1 = 2, whereas the solutions from case 2 are (x, y) = x2 + y 2 = 3 + 3 = 6. The two solutions from case 1 are therefore minima whereas the two solutions from case 2 are maxima. 5