ANSWERS. Part 1 2. i) 1000 ii) iii) iv) 501 v) x x a) i) 4 ii) 3,4 b) p=10,9

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1 ANSWERS. Part. i) 000 ii) 000 iii) 50 iv) 50 v) x +0x+0.. a) i) ii), b) p=0,9. a) i) 0 ii) 9,09 iii) 00 b) The INCREASE in cost incurred when you clean the lake above 50%, i.e. the marginal cost of further reduction after pollution has been decreased by 50%. 5. i) ii) 5 iii) /8 6. i) /5 ii) a,5 iii) 9 aa7 7. a) K= 5*L / b) x= -b/a 8. i) 50* ( + p 00 ) ii) 50* ( + p 00 ) iv) 50* ( + p 0 ) 00 iii) 50* ( + p 00 ) q 9. a) (notice the direction) b) c) d) neither e) 0. No. Not if x is negative. b) Yes. a) ln b) 0,5ln c)0, ln d) - ln. a) ln 8 ln = b) e ln ln a) tt = llllll bb aa b) llll aa c) d) ln ln8 ln ln e) e 6/7 f) ln ln 5. a) True b) False c) True d)false e) False 7. a) per cent of the male wage and 5 per cent of the female. b) 0.6 percentage points which is 7. per cent of male unemployment and 7.8 per cent of female unemployment % 9. a) They increase by approximately % b) They decrease by 5 %, (Hint: Calculate the new wage and price level.)

2 Part.. i-vi are all linear functions except iv. Intercept Slope i. ii. -/ iii 0,5 - v 0 vi 5 0,5. i) y = x + ii. y = -.5x iii. y = 0.00x.5. a) x = and y = - b) x = and y = c) x = ; y = -5 and z =. P(, ), L=x-, M=,5-0,75x a) kk GG =. A one unit increase in public consumption increases GNP by k G units, ( bb rr) which is more than one. b) kk TT = bb. Thus, k T = -bk G. Since b <, k T < k G. If G and T are increased by the same ( bb rr) amount, the result is a net increase in Y. c) kk GG = Since (-t) <, ( b(-t)) > (-b) and the multiplicator is smaller than ( bb( tt) rr) in a) since some of the additional demand created by an increase in G is siphoned off by taxes. 6. a) No solution. x 0 for any real number x so x cannot be equal to - b) Two solutions, x = and x = - c) One solution, x = 0 d) Infinitely many solutions e) One solution, x = and y =. (x-) 0 and (y-) 0. The sum of two non-negative numbers can be = 0 if and only if they are both equal to zero. 7. a) K = 000; L = 000 b) x = 6, y = 6 c) x = 6; y = Part.. i) f(x)=(x+)(x+) so x=- and x=- are roots ii) f(x )= /6*(x+8)(x-) so x=-8 and x= are roots iii) f(x)= =(x-) so x = is a double root iv) no real roots. a) x = 0; x = ; x = - b) p(x) = (x-)(x + 7/ x -) so the zeroes are x = / ; x = ; x = - Part (note that many of the answers can be simplified.) Part.

3 . a) 0 b) x c) 90x 9 d) 0 (π is a constant number).. a) -g (x)/6 b) g (x)/. a) x b) x - c) x -,5 d)-,5a -.5 e) +[a+f (a)]/h a) b) +x c) 5x +8x d) x +x -/ e) ½ - x + 5x 5. a) x 5 -x -x b) 0t 9 +5t +t -t - c),5x 0,5-0,5x -,5 6. xx( xx+) 7. SS (pp) = pp dddd(pp) dddd Part. and xx = 5. a) x= b) x= ±, x = 0 c) x =0, xx = 5. a) y=-x b) y=x- c) yy = 7 9 xx =.5xx.75 d) x 9 ad bc (aaaa+bb). a) b) ( ct + d) (hint: yy(tt) = (aatt +bbbb+cc) (aatt + bbbb + cc) ) c) ( y + 8 ) 9 y 5. a) 6(x+) b) -5(-x) c) -(x+)(x +x-) - xx d) (Hint: FF = (xx + ) 0.5 ) xx + e) f) -66x(-x ) g) -6at(at +) - h) an(at+b) n- ( x ) / ( x + ) / 6. a) f (x)+ b) f(x) f (x) c) f(x)+xf (x) d) 0,5f(x) -0,5 f (x). 7. a) 0x(+x ) t 8. 5 t ( + ) 9. Aab(bt+c) a- 5( x + ) 0. a) b) 0,5(x+(x+x 0,5 ) 0,5 ) -0,5 (+0,5(x+x 0,5 ) -0,5 (+0,5x -0,5 )) ( x + x + ) 6 c) ax a- (px+q) b + x a bp(px+q) b-. a) aa (tt) aa(tt) + bb (tt) bb(tt) b) La ( t) + a( t) Bb ( t) b( t) LLLL (tt) aa(tt) + BBBB (tt) bb(tt). a) q(5-0,5x) -/ b) f (x n g(x))(nx n- g(x)+x n g (x)). a) 0x -6x b) -0,5x -,5 c) 0(+x ) 8 (+9x ). a) - b) g () (u(x))(u (x)) +g (u(x))u () (x) 5. a) 80 b)

4 Part. x. a) -e -x b) 6x exp(x ) c) e d) 5(x-)exp(x -x+) x. a) exp(x)*exp(exp(x )). b) 0,5(e t/ -e -t/ x x e e ) c) x x ( e + e ) d) z exp( z ) (exp( z ) ) / e) e x (+x). a) (x+) - b) x - c) lnx+ d) ln x (ln x) g) ex lnx+x - e x x + e) x + x.. a) x- b) e - x 5. a) x- b) e - x / ( x + ) 6. a) ( x )( x ) / b) x x (lnx+) c) + ( x ) x x x + + * y. x + 6 Part.. a) - b) 00 c) 0,5 d)-,5 e) 0 f) xx xx+ g) -0x (-x ) - h) xg ( x ) i) (+x)(+x) - j) bx b (a+x b ) - k) -b(ax+b) - g( x) Part Optimization (one independent variable). a) min (0, ); max (-, 6) b) min (-, -0,5); max (, 0,5) c) min (, 6 ); max (, 6 ) ; inflection point (0, 0) e) min ( /8, -6); max ( -/, 0), inflection point (, 0). a) Hint: Add and subtract to the numerator, then write as a sum of two quotients.. min (7, 0); max(, 6). a) L= (AK α ) -/β Q /β. This is the equation for the isoquant for production level Q (where Q is a constant). It shows the combinations of capital (K) and labour (L) required to produce Q units. 5 a) max point at (,5;,97) b) min point at (, ); max point at (, ) 6. a) V(x)= x(8-x) b) x= maximizes V(x). c) V(x) is strictly increasing when x< and x>9. V(x) is strictly decreasing <x<9. 7. b) No solution.

5 8. a) π= 70Q-Q -900 b) Q 6,97 or Q 5 c) Q=5. 9 a) No stationary points. b) min point at (0, 0); max point at (, e - ) 5

6 Part 5 Functions of more than one variable.. a) i) 0 ii)- iii) a b) i) ii) 5. a) 5/6 0K / L / b) α=5/6, The production function exhibits decreasing returns to scale.( α=5/6<) dz dz dk dk = x; = 6y = y; = x. a) dx dy b) dx dy dp 5 dp = 0x y y ; = 0x y 0xy c) dx dy df df df = yz + xy - z ; = xz + x ; = xy - xz. a) dx dy dz dy dy KL (al bk ) b) = ; dl = LK (bk al ) (al + bk ) dk (al + bk ) c) TT xx = xxxx xx yy yy yy +xx xx xx yy yy = 8xxyy xx (yy xx +6xx ) (yy +xx ) (yy +xx ) TT yy = (xx yy yy)(xx )(yy + xx ) 8yy (xx yy yy) (yy + xx ) If you use logarithmic differentiation you get the same answers in this form: x x T x= x (x y y) ; T y= 8y (x y y) y + x (y + x ) y y + x (y + x ) 5. a) dz = (x-8y)dx (8x+y )dy b) total derivative = z (t) = t + 60 t ; total differential dz = (t + 60 t )dt c) z(t) = t +0t -t-. a) 00 b) dq =,5K -/ L / dk + 5K / L -/ dl. With K=65; L = 6; dk=0.05; dl=0. Q(6.05; 65.) = 00,75 Analogously, Q(00, 675) 58.5 c) Q(6.05; 65.) 00.7 Q(00; 675) ) 58,

7 Part 6 Optimisation of functions of several variables Part 6. Unconstrained optimisation. a) The only stationary point is (, ) b) The stationary point is (, 9). a) f(, -)=0 c) (, -) is a global minimum point of f. L= 5,75; K=,8. a) min point f(, )= -; saddle point f(0, 0)=0 b) max point f(, )= c) min points at f(0,5, 0,5) and f(-0,5-0,5)= -0,065; saddle at f(0, 0)=0. d) we have no way of knowing using this method. 5. a= ; b= -; c= /7. Part 6. Constrained optimisation. f(/, /) = 5/ c) λ= implies that the rate of change in the optimal value of the objective function f(x,y) when the constraint is slightly increased is one.. a) Stationary points at (, ), (-, -), (-, ) and (, -). Minimum values at f(-, ) and f(, -)=-, Maximum values at f(, ) and f(-,-)=. (x +y =8 is a circle with its centre at (0; 0) and radius 8. The points for which f(x, y) = k are on a graph y = /x which has a similar shape to y= /x. For each k>0, it has one branch in the first quadrant and one in the third, for each k<0 it has one branch in the second and one in the fourth quadrant. The larger k is, the further from origin is y = k/x.). (0.a 0.a) is a min point.. U(0, 60)= 0*0 / 60 / 5. xx = RR qq(llllll llllll) ; yy = RR pp(llllll llllll) (pp+qq) (pp+qq) 6. Min cost=50 at (0, 0) 7. Min cost 80,5 when K=80* / and L= 80* -/ 8. max at F(, )=ln 8. (Hint: Use the rules of logarithms to simplify or find the point that maximises G(x, y) = xy instead.) 7