<p>E604 Suggested Answers to Problem Set #2 (Problems 2.3, 2.5, 2.6, 2.9)</p><p>2.3. The height of a ball t seconds after it is thrown straight up is -1/2gt2 + 40t (where g is the acceleration due to gravity) a. If g=32 (as on the earth), when does the ball reach a maximum height? What is that height? h = -1/2gt2 + 40t dh/dt = -gt +40 = 0</p><p>Given g= 32, t = 40/32 = 1.25 seconds h = 25 feet b., If g=5.5 (as on the moon) when does the ball reach a maximum height and what is that height? Can you explain the reasons for the difference between this answer and the answer for part (a)? dh/dt = -gt +40 = 0</p><p>Given g= 5.5, t = 40/5.5 = 7.27 seconds h = 145.45 feet. </p><p>The weaker gravity of the moon damps the upward thrust of the ball more slowly c. In general, develop an expression for the change in maximum height for a unit change in g. Explain why this value depends implicitly on the value of g itself. </p><p>Via the envelope theorem. h* = -1/2g t*2 + 40(t*) thus dh*/dg = -1/2t*2</p><p>Note from the FONC t*(g) = 40/g. Thus dh*/dg = -800/g2. The optimal h depends on the value of g because t* is a function of g. (Note that you could have alternatively worked this problem by inserting the relationship between g and optimal t derived from FONC, inserted this into the objective function and then took the derivative. 2.5 Suppose U(x,y) = 4x2 + 3y2 </p><p> a. Calculate U/x, U/y</p><p>U/x = 8x U/y = 6y</p><p> b. Evaluate these partial derivatives at x=1, y=2</p><p>U/x(x=1) = 8 U/y (y=2) = 12</p><p> c. Write the total differential for U dU = U/x dx + U/y dy = 0 8x dx + 6y dy</p><p> d. Calculate dy/dx for dU=0 – that is, what is the implied trade-off between x and y holding U constant? dy/dx = -8x/6y = -4x/3y. </p><p>The negative sign indicates that getting more x requires forgoing some y and vice versa. Further the tradeoff is not linear. Starting at any given values for x and y, an increment of y requires foregoing 4x/3y of x. The size of what is foregone will change with x and y. However, to asses the rate of tradeoff, we must express the slope in terms of a single variable (see g below)</p><p> e. Show U =16 when x=1, y=2</p><p>Inserting into the objective function</p><p>U = 4(1) 2 + 3(2)2 = 16</p><p> f. In what ratio must x and y change to hold U constant at 16 for movements away from x=1, y=2? dy/dx = -8(1)/6(2) = -8/12 = -2/3</p><p> g. More generally, what is the shape of U=16 contour line for this function? What is the slope of that line?</p><p>Solving the objective for y when U=16 yields y = (2/3)(4-x2)1/2 2.5</p><p>2</p><p>1.5 Y</p><p>1</p><p>0.5</p><p>0 0 0.5 1 1.5 2 2.5 X</p><p>The slope of that line is -2x/[3)(4-x2)1/2]</p><p>2.6 Suppose that f(x,y) = xy. Find the maximum value for f if x and y are constrained to sum to 1. Solve this problem in two ways: by substitution and by using the Langragian mulitipler method. </p><p> a. Substitution method f(x,y) = xy with substitution f(x,y) = x(1-x) fx = 1 – 2x x = ½</p><p> b. Lagrangian method. Write the constraint in implicit function form: 1-x-y = 0</p><p>L = xy - (1-x-y)</p><p>Taking first order conditions</p><p> L/x =y -  = 0</p><p> L/y =x -  = 0</p><p>L/ = 1 - x-y = 0</p><p>Solving x = y Thus, 1-2x = 0 x =1/2</p><p>2.9. One of the most important functions we will encounter in this book is the Cobb- Douglas function:   y = x1 x2 where  and  are positive constants that are each less than one. a. Show that this function is quasi-concave using a “brute force” method by applying Equation 2.107</p><p>-1  f1 = x1 x2  -1 f2 = x1 x2</p><p>-2  f11 = (-1)x1 x2  -2 f22 = (-1)x1 x2 -1 -1 f12 = ()x1 x2</p><p>Thus, by 2.107</p><p>2 2 f11 f2 - 2 f12 f1 f2 + f22 f1 -2   -1 2 -1 -1 -1   -1  -2( -1  2 (-1)x1 x2 (x1 x2 ) - 2()x1 x2 x1 x2 x1 x2 +(-1)x1 x2 x1 x2 ) 2 3-2 3-2 2 2 3-2 3-2 2 3-2 3-2 (-1) x1 x2 - 2  x1 x2 +(-1) x1 x2 2 2 2 2 3-2 3-2 [(-1) + (-1) - 2  ] x1 x2</p><p>Notice that with 0<<1 and 0<<1 each of the terms in the final expression are negative. QED b. Show that the Cobb-Douglas function is quasi-concave by showing that any contour line of the form y=c (where c is any positive constant) is convex and therefore that the set of points for which y>c is a convex set.</p><p>  c = x1 x2 implies that -/ x1 = Kx2 where K = c1/. (--)/ dx1/dx2= (-/)x2 <0 2 2 2 (--2)/ d x1/dx2 = [-(--)/ ]x2 >0</p><p>Thus, for a “level curve” defined by a given value c, the slope between x and y is negative (the first order condition) and it gets negative at a diminishing rate (the second order condition). The tradeoff between the two variables is convex. Thus the function itself is quasi-concave. c. Show that if  +  >1 then the Cobb-Douglas function is not concave (thereby illustrating that not all quasi-concave functions are concave). 2 Concavity requires that f11f22 – f12 >0 and f11<0, f22<0 From the second order conditions in part (a) -2  f11 = (-1)x1 x2  -2 f22 = (-1)x1 x2 -1 -1 f12 = ()x1 x2</p><p>2 2-2 2-2 2 2 2-2 2-2 Thus f11f22 –f12 =  (-1)(-1) x1 x2 -   x1 x2 2-2 2-2 =  (1--) x1 x2 <0 for +>1 </p>