Chapter 4 Differentiation 08 Section 4. The derivative of a function Practice Problems (a) (b) (c) 3 8 3 ( ) 4 3 5 4 ( ) 5 3 3 0 0 49 ( ) 50 Using a calculator, the values of the cube function, correct to decimal places, are x.50.5.00 0.75 f( x) 3.38.95.00 0.4 x 0.50 0.5 0.00 0.5 0.50 f( x) 0.3 0.0 0.00 0.0 0.3 x 0.75.00.5.50 f( x) 0.4.00.95 3.38 The graph of the cube function is sketched in Figure S4..
09 Figure S4. f '( ).5 0.5 3.0 f '(0) 0 (because the tangent is horizontal at x 0) f '().5 0.5 3.0 [Note: f'( ) f'() because of the symmetry of the graph.] 3 If n 3 then the general formula gives Hence f '(x) 3x 3 3x f '( ) 3( ) 3 f '(0) 3(0) 0 f '() 3() 3
4 (a) 5x 4 ; (b) 6x 5 ; (c) 00x 99 ; 0 (d) x (that is, /x ); (e) x 3 (that is, /x 3 ). Exercise 4. (p. 49) (a) 9 5 ; (b) 4 5 ( ) ; (c) 7 3 9 9 0. 4 7 0+ 3 6 and 3+ 3 0 6 0 3 0 3 which is negative so the line is downhill. 3 The graph of f(x) 5 is sketched in Figure S4.. The graph is horizontal, so has zero slope at all values of x. Figure S4. 4 7x 6 ; 6 f '() 7 448 5 (a) 8x 7 ; (b) 50x 49 ; (c) 9x 8 ; (d) 999x 998. 6 (a) (b) f ( x) x f '( x) 3x 3 x 3 4 4 f( x) x f '( x) x x
(c) 3 f( x) x x 3 x x x (d) 3 3 3 x f ( x) x x x f '( x) x 7 3,.5, 0, 0.75,, 0.75, 0,.5; (a) 3; (b) 0; (c). Section 4. Rules of differentiation Practice Problems (a) 4(3x ) x. (b) /x 3 because /x x, which differentiates to x. (a) 5x 4 + ; (b) x + 0 x. 3 (a) x 3x ; (b) 0 ( 3x 4 3 ) 4 x. 4 (a) 9(5x 4 ) + (x) 45x 4 + 4x (b) 5(8x 7 ) 3( )x 40x 7 + 3/x (c) x + 6() + 0 x + 6 (d) (4x 3 ) + (3x ) 4(x) + 7() 0 8x 3 + 36x 8x + 7 5 f '(x) 4(3x ) 5(x) x 0x f "(x) (x) 0() 4x 0 f "(6) 4(6) 0 34
Exercise 4. (p. 58) (a) 0x; (b) 3/x ; (c) ; (d) x + ; (e) x 3; (f) 3 + 7/x ; (g) 6x x + 49; (h) a; (i) ax + b; (j) (a) 3 3 4 f( x) 4x 3x + 7 x f '( x) x + 3x 4x + 3 x x x f x x f 8 '( ) 7 '() 7 (b) f '( x) x f '(3) 4 (c) (d) (e) f x x x f '( ) 3 8 + '(0) f x x x f x x x f 4 4 3 5 ( ) 5 4 '( ) 0 + 6 '( ) 36 3 f( x) x x f '( x) x + x f '( x) 8 3 4x 3 + 6x; (a) (b) (c) (d) (e) (f) 4 (a) (b) 3 f ( x) 3x 4 x f '( x) 9x 8x f x x x x x f x x x x 4 3 3 ( ) 3 + 6 7 '( ) 6 + 7 f x x x f x x ( ) 5 6 '( ) 5 3 f ( x) x 3 x f '( x) + 3x + x f( x) x 4 x f '( x) x 4x 3 + + 3 f ( x) 3x 5 x f '( x) 3x 0x 4 x x 3 0 x x 3 + 3 dy d y 4x 4 dx dx dy d y 6 y x x x dx dx x 3 4 6 4
(c) dy a dx dx d y 0 3 5 f x x x f x x f '( ) 3 8 + 0 "( ) 6 8 "() 4 6 f '( x) x 6 f '(3) 0 ; horizontal tangent, i.e. vertex of parabola must be at x 3. 7 f( x) x f '( x) x x (a) (b) (c) 5 5 f( x) 5x 5 x 5 x f '( x) x x 3 3 3 3 3 f ( x) 7x 7 x 3 x f '( x) x 3 4 3 4 4 3 3 3 4 4 f( x) 6x 6 x x f '( x) x 4 x 5 5 5 (d) f ( x) 5 x f '( x) x x x 8 (a) P + (b) 50 6Q; 3 (c) d( AC) 30 AC 30Q + 0 30Q dq Q (d) 3; (e) dq 5 Q 0L 5L dl L (f) 6Q + 30Q 4. Section 4.3 Marginal functions Practice Problems TR PQ (60 Q)Q 60Q Q
() MR 60 Q 4 When Q 50, MR 60 (50) 40 () (a) TR 60(50) (50) 500 (b) TR 60(5) (5) 459 so TR changes by 4, which is approximately the same as the exact value obtained in part (). MR 000 8Q, so when Q 30 MR 000 8(30) 760 (a) Δ(TR) MR ΔQ 760 3 80, so total revenue rises by about 80. (b) Δ(TR) MR ΔQ 760 ( ) 50 so total revenue falls by about 50. 3 TC (AC)Q 00 + Q 00 + Q Q This function differentiates to give MC, so a unit increase in Q always leads to a unit increase in TC irrespective of the level of output. 4 If K 00 then Q 5L / (00) / 50L / because 00 0. Differentiating gives MP L 50( /L / ) 5 L (a) (c) 5 5 (b) 5 5 8.3; 9 3 5 5 0.5. 0000 00
5 The fact that these values decrease as L increases suggests that the law of diminishing marginal productivity holds for this function. This can be confirmed by differentiating a second time to get dq 3/ 5 5( / L ) 3/ dl L which is negative for all values of L. 5 The savings function is given, so we begin by finding MPS. Differentiating S with respect to Y gives MPS 0.04Y so when Y 40, MPS 0.04(40) 0.6 To find MPC we use the formula MPC + MPS that is, MPC MPS 0.6 0.4 This indicates that, at the current level of income, a unit increase in national income causes a rise of about 0.6 units in savings and 0.4 units in consumption. Exercise 4.3 (p. 73) TR PQ (00 4 Q) Q 00Q 4Q dtr ( ) MR 00 8Q dq When Q, MR 4 so Δ( TR) 4 0.3. TR PQ (80 3 Q) Q 80Q 3Q, so MR dtr ( ) dq 80 6Q 80 6(80 P)/3 P 80.
3 TR PQ (00 Q) Q 00Q Q dtr ( ) MR 00 Q dq Graphs of TR and MR are sketched in Figures S4.3 and S4.4 respectively. MR 0 when Q 50. This is the value of Q at which TR is a maximum. 6 Figure S4.3 4 Figure S4.4 5 TC AC Q Q 9 Q 5 Q + + + + 9Q Q Fixed costs are 5 since this is the constant term in the expression for TC dtc ( ) MC 4Q + 9 dq
dq 5 MPL 50 0.0L dl 7 (a) 49.98; (b) 49.8; (c) 48; (d) 30. Yes, because d Q/dL 0.0 < 0. 6 dc C 50 + Y 50 + Y MPC Y dy When Y 36, MPC 6 so 5 MPS MPC 6 If national income rises by unit, the approximate increase in consumption and savings is /6 and 5/6 respectively. dc 7 MPC 0.04Y + 0. dy MPS MPC (0.04Y + 0.) 0.9 0.04Y 0.9 0.04Y 0.38 0.04Y 0.5 Y 3 Section 4.4 Further rules of differentiation Practice Problems (a) The outer power function differentiates to get 5(3x 4) 4 and the derivative of the inner function, 3x 4, is 3, so dy dx 5(3x 4)4 (3) 5(3x 4) 4
8 (b) The outer power function differentiates to get 3(x + 3x + 5) and the derivative of the inner function, x + 3x + 5, is x + 3, so dy dx 3(x + 3x + 5) (x + 3) (c) Note that y (x 3). The outer power function differentiates to get (x 3) and the derivative of the inner function, x 3, is, so dy dx (x 3) () (x 3) (d) Note that y (4x 3) /. The outer power function differentiates to get /(4x 3) / and the derivative of the inner function, 4x 3, is 4, so dy dx /(4x 3) / (4) (4x 3) (a) u x v (3x ) 6 du dx dv dx 6(3x )5 So dy dx 8x(3x )5 + (3x ) 6 (3x ) 5 [8x + (3x )] (3x ) 5 (x ) (b) u x 3 v (x + 3) / du dx 3x dv dx /(x + 3) / (x + 3)
So 9 3 dy x + + dx (x + 3) (c) u x v (x ) 3 x (x 3) du dx dv dx (x ) So dy x + dx x x ( ) x+ ( x ) ( x ) ( x ) 3 (a) u x v x dy dx dv dx So dy ( x ) x dx ( x ) ( x ) (b) u x v x + du dx dv dx So dy ( x + ) ( x ) dx ( x + ) ( x + )
0 Exercise 4.4 (p. 83) (a) (b) (c) (d) (e) (f) (g) (h) (i) (a) (b) (c) (d) (e) dy 3(5x ) 5 5(5x ) dx + + dy 8(x 7) 6(x 7) dx 7 7 dy dx + + 4 4 5( x 9) 5( x 9) ; dy dx 3(4x 7) 8x 4 x(4x 7) ; dy dx + + + + 4( x 4x 3) (x 4) 8( x )( x 4x 3); dy y (x+ ) (x+ ) ; dx x + dy 3 y (3x+ ) ( )(3x+ ) 3 dx (3x + ) dy 3 8 (4 3) ( )(4 3) 4 ; 3 y x x dx (4x 3) 3 3 dy y (x+ 5) (x+ 5) (x+ 3) dx (x+ 3) (x+ 3) (3x+ 4) + x (3x+ 4) (3) (3x+ 4) + 6 x(3x+ 4) (9x+ 4)(3x+ 4); 3 3 x ( x ) + x 3( x ) () x( x ) + 3 x ( x ) x(5x 4)( x ) ; dy x 3x + 4 y x( x+ ) ( x+ ) + x ( x+ ) () x+ + dx x+ x+ 3 3 ( x+ 6) + ( x ) 3( x+ 6) () ( x+ 6) + 3( x )( x+ 6) (4x+ 3)( x+ 6) ; 3 3 ( x+ 5) + (x+ ) 3( x+ 5) () ( x+ 5) + 3(x+ )( x+ 5) (8x+ 3)( x+ 5) ;
(f) 4 3 3 4 3 3 3 3 x (x 5) + x 4(x 5) () 3 x (x 5) + 8 x (x 5) x (4x 5)(x 5). ( x 5) x x 5 x 3 (a) ( x 5) ( x 5) 5 ( x 5) ; ( x+ 7) x x+ 7 x 7 ( x+ 7) ( x+ 7) ( x+ 7) (b) ( x ) ( x+ 3) x x 3 5 ( x ) ( x ) ( x ) (c) (3x+ ) (x+ 9) 3 6x+ 6x 7 5 (3x+ ) (3x+ ) (3x+ ) (d) (5x+ 6) x 5 5x+ 6 5x 6 (5x+ 6) (5x+ 6) (5x+ 6) (e) (3x 7) ( x+ 4) 3 3x 7 3x 9 (3x 7) (3x 7) (3x 7) (f) 4 (a) (5x+ 7)(5) 0(5x+ 7) ; ; (b) 5 (a) (b) dy y x + x+ x+ x+ dx 5 70 49 50 70 0(5 7) 4 5 4 5 5 x ( x+ ) + x ( x+ )() 5 x ( x+ ) + x ( x+ ) 5 x ( x + 4x+ 4) + x + 4x 7x + 4x + 0x 4 6 5 6 5 4 dy y x ( x + 4x+ 4) x + 4x + 4x 7x + 4x + 0x dx 5 7 6 5 6 5 4 6 (a) 3 TR PQ (00 Q) Q dtr ( ) MR 3(00 Q) ( ) Q+ (00 Q) 3 Q(00 Q) + (00 Q) dq 3 3 (00 4 Q)(00 Q) ;
(b) 000Q TR PQ Q + 4 dtr ( ) ( Q+ 4) 000 000Q 4000 MR dq ( Q + 4) ( Q + 4) 7 ( + ) 4 (300 + ) + 4 300 dc Y Y Y Y Y MPC dy ( + Y ) ( + Y ) When Y 36, MPC.78 and MPS MPC 0.78. If national income rises by unit, consumption rises by.78 units, whereas savings actually fall by 0.78 units. Section 4.5 Elasticity Practice Problems We are given that P 0 and P 00. Substituting P 0 into the demand equation gives 000 Q 0 Q 790 Q 395 Similarly, putting P 00 gives Q 400. Hence Δ P 00 0 0 Δ Q 400 395 5 Averaging the P values gives P (0 + 00) 05
Averaging the Q values gives 3 Q (395 + 400) 397.5 Hence, arc elasticity is 05 5 0.6 397.5 0 The quickest way of solving this problem is to find a general expression for E in terms of P and then just to replace P by 0, 50 and 90 in turn. The equation P 00 Q rearranges as so Q 00 P Hence dq dp P dq P E ( ) Q dp 00 P P 00 P (a) If P 0 then E /9 < so inelastic. (b) If P 50 then E so unit elastic. (c) If P 90 then E 9 so elastic. At the end of Section 4.5 it is shown quite generally that the price elasticity of demand for a linear function P aq+ b is given by P E b P
The above is a special case of this with b 00. 4 3 Substituting Q 4 into the demand equation gives P (4) 0(4) + 50 94 Differentiating the demand equation with respect to Q gives so dp Q 0 dq dq dp Q 0 When Q 4 dq dp 8 The price elasticity of demand is then 94 47 4 8 36 From the definition E we have percentage change in demand percentage change in price 47 0 36 percentage change in price Hence the percentage change in price is 0 36/47 7.7%: that is, the firm must reduce prices by 7.7% to achieve a 0% increase in demand. 4 (a) Putting P 9 and directly into the supply equation gives Q 03. and 7. respectively, so Δ P 9 Δ Q 7. 03. 4 Averaging the P values gives
P (9 + ) 0 5 Averaging the Q values gives Q (03.+ 7.) 0. Arc elasticity is 0 4 0.33375 0. (b) Putting P 0 directly into the supply equation, we get Q 0. Differentiating the supply equation immediately gives dq 5 0.P dp + so when P 0, dq/dp 7. Hence 0 E 7 0 3 Note that, as expected, the results in parts (a) and (b) are similar. They are not identical, because in part (a) the elasticity is averaged over the arc from P 9 to P, whereas in part (b) the elasticity is evaluated exactly at the midpoint, P 0. Exercise 4.5 (p. 98) When Q 8, P 500 4 8 44 When Q 0, P 500 4 0 00 Δ P 00 44 44 Δ Q 0 8 P ( 44 + 00 ) 7 Q (8 + 0) 9 Hence P ΔQ 7 43 E 0.7 (to decimal places) Q ΔP 9 44 6
Putting Q 9 into the demand function gives P 500 4 9 76 6 dp dp 8Q so when Q 9, 7 dq dq Hence dq dp 7 P dq 76 E 0.7 Q dp 9 7 8 (to decimal places) The values agree to decimal places. 3 (a) 30 Q 6 Q 4 Q dp dq so dp dq Hence P dq 6 E Q dp 4 (b) 30 Q 6 Q 4 Q dp dq so dp dq Hence P dq 6 E Q dp 4
(c) 7 00 Q 6 00 Q 36 Q 64 Q 3 dp P Q Q Q dq ( 00 ) ( 00 ) ( ) ( 00 ) so Hence dp dq giving 6 dq 6 dp P dq 6 9 E ( 6) Q dp 3 8 4 (a) 0.P. (b) 0.P Q 4 P Q Q 0( 4) 0 40 P (0Q 40) dp 5 P (0Q 40) (0Q 40) (0) dq (0Q 40) (c) (0Q 40) d P/dQ 5 P dq 0.P 5 dp dq (d) Q 4 P 0 and dp P dq 0 0 E Q dp 4 7
dq 5 0. 0.008P dp + 8 dq At P 80, Q 40.6 and 0.74 dp Hence P dq 80 E 0.74.46 Q dp 40.6 (a) elastic; (b) % change in supply 5.46 7.3 %. Section 4.6 Optimization of economic functions Practice Problems (a) Step dy 6x 0 dx + has solution x. Step d y d 6 0 x > so minimum. Finally, note that when x, y 47, so the minimum point has coordinates (, 47). A graph is sketched in Figure S4.5.
9 Figure S4.5 (b) Step dy x x dx + 6 30 36 0 has solutions x and x 3. Step d y d x x 30 which takes the values 6 and 6 at x and x 3 respectively. Hence minimum at x and maximum at x 3. Figure S4.6
A graph is sketched in Figure S4.6 based on the following table of function values: 30 X 0 0 3 0 f(x) 3887 7 0 833 4 Q 300L L APL 300L L L L 3 Step d(ap L ) dl L 300 3 0 has solution L ±0. We can ignore 0 because it does not make sense to employ a negative number of workers. Step d(ap) L dl 6L which takes the value 60 < 0 at L 0. Hence L 0 is a maximum. Now dq MPL 600L 4L dl so at L 0 3 MP 600(0) L 4(0) 000 3 AP L 300(0) (0) 000 that is, MP L AP L. 3 (a) TR PQ (0 Q)Q 0Q Q Step d(tr) 0 4Q 0 dq has solution Q 5. 3
Step 3 d(tr) 0 dq < so maximum. (b) π TR TC + + 3 (0Q Q ) ( Q 8Q 0Q ) Q + Q Step 6 dπ dq + 3Q Q 0 has solutions Q 0 and Q 4. Step d π Q dq + 6 which takes the values and when Q 0 and Q 4, respectively. Hence minimum at Q 0 and maximum at Q 4. Finally, evaluating π at Q 4 gives the maximum profit, π 30. Now d(tr) MR 0 4Q dq so at Q 4, MR 4; d(tc) dq + MC 3Q 6Q 0 so at Q 4, MC 4.
4 36 AC Q + 3 + Q Step 3 d(ac) 36 0 dq Q has solution Q ±6. A negative value of Q does not make sense, so we just take Q 6. Step d(ac) 7 dq Q 3 is positive when Q 6, so it is a minimum. Now when Q 6, AC 5. Also d(tc) MC Q 3 dq + which takes the value 5 at Q 6. We observe that the values of AC and MC are the same: that is, at the point of minimum average cost average cost marginal cost There is nothing special about this example and in the next section we show that this result is true for any average cost function. 5 After tax the supply equation becomes P Q + + t S 5 In equilibrium, Q S Q D Q, so Hence P Q+ 5 + t P Q+ 50 Q+ 5 + t Q+ 50 which rearranges to give
Q 0 t 5 33 Hence the tax revenue, T, is T tq 0t t 5 Step dt 4 t dt 5 0 0 has solution t.5. Step d T 4 dt 5 < 0 so maximum. Government should therefore impose a tax of $.50 per good. Exercise 4.6 (p. 36) dy (a) Step x dx + At a stationary point, x + 0 x x 5 The coordinates of the stationary point are, 4 Step d y < 0 so the point is a maximum. dx
The graph is sketched in Figure S4.7. 34 Figure S4.7 dy (b) Step x 4 dx At a stationary point, x 4 0 x 4 x The coordinates of the stationary point are (,0 ) Step d y > 0 so the point is a minimum. dx The graph is sketched in Figure S4.8. Figure S4.8 dy (c) Step x 0 dx At a stationary point,
x 0 0 x 0 x 0 35 The coordinates of the stationary point are ( 0,5 ) Step d y > 0 so the point is a minimum. dx The graph is sketched in Figure S4.9. Figure S4.9 (d) Step dy dx + 3x 3 At a stationary point, + 3x 3 x x ± 3x 3 0 The coordinates of the stationary points are (, ) and (,)
Step d y 6 dx x 36 At x, At x, d y 6> 0 so this point is a minimum dx d y 6< 0 so this point is a maximum dx Maximum at (, ), minimum at (, ) The graph is sketched in Figure S4.0. Figure S4.0 TR PQ (40 Q) Q 40Q Q Step dtr ( ) 40 4Q dq At a stationary point, 40 4Q 0 4Q 40 Q 0 Step d ( TR) 4< 0 so the stationary point is a maximum. dq
3 Q APL 30L 0.5L L 37 Step Step d( AP L ) 30 L so there is one stationary point at L 30. dl d ( AP L ) < 0so it is a maximum. dl At L 30. AP 30 30 0.5 30 450 dq MPL 60L.5L dl L At L 30, MP 60 30.5 30 450 4 (a) TC 3 + ( Q+ ) Q L + + 3 Q Q TC 3 AC + Q + Q Q Q 3 4 5 6 AC 6 0.5 9.3 9.3 9.6 0.
The graph of AC is sketched in Figure S4.. 38 Figure S4. (b) From Figure S4. minimum average cost is 9.. (c) Step d( AC) 3 3 3 AC Q + Q Q + + dq Q At a stationary point 3 + 0 Q Q 3 Q 3 Step d ( AC) 3 6Q > 0 when Q 3 so the stationary point is a minimum. dl At Q 3, AC 9..
5 (a) Q Q TR PQ 4 Q 4Q 4 4 39 3 3 Q 3Q Q Q Q π TR TC 4Q 4+ Q + + + Q 4 4 0 0 0 0 d(tr) Q MR 4 dq d(tc) 3Q 3Q MC + dq 5 0 (b) Step d Q Q π 3 + + Q Q dq 0 0 0 3 40 0 ± ± Q (3) 6 ( ) ( ) 4(3)( 40) 484 ± 0 Q, 4 6 3 Ignoring the negative solution gives, Q 4. Step d π 3Q + dq 0 0 At Q 4, d π < 0 so maximum. dq 0 (c) dtr ( ) Q MR 4 which takes the value at Q 4 dq ( ) 6 3 MC dtc Q + Q also takes the value at Q 4. dq 0 0 6 The new supply equation is 3( P t) Q 3 3P 3t Q 3 S S which can be reaaranged as Q 3P 3t 3 S
The demand equation can be rearranged as Q 4 P. D 40 In equilibrium, QS QD Q so 3P 3t 3 4 P 5P 3t 3 4 5P 3t+ 7 P (3 t+ 7) 5 Hence 6 Q 4 (3t+ 7) (6 t) 5 5 6 6 TR PQ t + t t + t + 5 5 (3 7)(6 ) ( 3 0) Step Step dtr ( ) 6 ( 6t+ ) 0 t dt 5 6 d ( TR) 36 < 0 so the stationary point is a maximum. dt 5 7 (a) TC Q Q + Q Q + ( 36) 00 36 00 TC 00 AC Q 36+ Q Q (b) Step d( AC) AC Q + Q Q + dq 00 36 00 At a stationary point, 00 0 00 0 Q Q + (ignoring the negative solution) Q Step d ( AC) 400Q dq 3 When Q 0, d ( AC) 0.4 > 0 so the stationary point is a minimum. dq
4 (c) Q 0 AC 4 dtc ( ) MC 4Q 36 dq Q 0 MC 4 AC Section 4.7 Further optimization of economic functions Practice Problems (a) TR (5 0.5 QQ ) 5Q 0.5Q TC 7 ( ) 7 + Q+ Q Q + Q+ MR 5 Q MC Q + (b) From Figure S4.3 the point of intersection of the MR and MC curves occurs at Q 8. The MC curve cuts the MR curve from below, so this must be a maximum point. Figure S4.3 MC 00. (a) Domestic market P 300 Q TR 300Q Q so MR 00 Q
To maximize profit, MR MC: that is, 4 300 Q 00 which has solution Q 00. Corresponding price is P 300 00 $00 Foreign market P 00 Q TR 00Q Q so MR 00 Q To maximize profit, MR MC: that is, 00 Q 00 which has solution Q 00. Corresponding price is P 00 (00) $50 (b) Without discrimination, P P P, say, so individual demand equations become Q 300 P Q 400 P Adding shows that the demand equation for combined market is Q 700 3P where Q Q + Q. 700 Q TR Q 3 3
so 43 700 MR 3 3 Q To maximize profit, MR MC: that is, 700 Q 00 3 3 which has solution Q 00. Corresponding price is P 700 / 3 00 / 3 $500 / 3 Total cost of producing 00 goods is 5000 + 00(00) $5000 With discrimination, total revenue is 00 00 + 00 50 $35000 so profit is $35000 $5000 $0000. Without discrimination, total revenue is 500 00 $33333 3 so profit is $33333 $5000 $8333. 3 Domestic market From Practice Problem, profit is maximum when P 00, Q 00. Also, since Q 300 P we have dq /dp. Hence P dq E Q dp 00 ( ) 00 Foreign market From Practice Problem, profit is maximum when P 50, Q 00. Also, since Q 400 P we have dq /dp. Hence
E P dq Q dp 44 50 ( ) 3 00 We see that the firm charges the higher price in the domestic market, which has the lower elasticity of demand. Section 4.8 The derivative of the exponential and natural logarithm functions Practice Problems x 0.50.00.50.00 f ( x) 0.69 0.00 0.4 0.69 x.50 3.00 3.50 4.00 f ( x) 0.9.0.5.39 The graph of the natural logarithm function is sketched in Figure S4.4. Figure S4.4 0.50 f ().00 0.50 0.5 f () 0.50 0.50 0.5 f (3) 0.30 0.50 3 These results suggest that f (x) /x.
(a) 3e 3x ; (b) e x ; (c) /x; (d) /x. 45 3 (a) For the product rule we put u x 4 and v ln x for which du 3 dv 4x and dx dx x By the product rule dy x + ln x 4x dx x 3 3 x + 4x lnx 4 3 3 x + ( 4 ln x) (b) By the chain rule (c) If then dy e x xe dx x x u ln x and v x+ du dv and dx x dx By the quotient rule ( x+ ) (ln x) dy x d x ( x+ ) x+ xlnx (multiply top and bottom by x) xx ( + ) 4 (a) y x x 3 4 ln + ln( + ) (rule ) 3 ln x+ 4ln( x+ ) (rule 3)
Hence 46 dy 3 4 + dx x x + 3( x+ ) + 4x 7x+ 6 xx ( + ) xx ( + ) (b) y x x ln ln( + 3) (rule ) ln x ln(x+ 3) (rule 3) Hence dy dx x x+ 3 (chain rule) (x + 3) x x(x+ 3) x + 6 x(x+ 3) 5 In terms of P the total revenue function is given by TR PQ 000Pe 0.P and the total cost function is TC 00 + Q 00 + 000e Hence π TR TC 0.P 0.P 0.P 000Pe 000e 00 Step At a stationary point dπ 0 dp To differentiate the first term, 000Pe 0.P, we use the product rule with u 000P and v e 0.P for which
du dv 000 and 0.e dp dp 0.P 47 Hence the derivative of 000Pe 0.P is dv du u + v dp dp + 0.P 0.P 000 P( 0.e ) e (000) 0.P e (000 00 P) Now π 000Pe 0.P 000e 0.P 00 so π 0.P 0.P e (000 00 ) 000( 0.e ) d P dp 0.P e (400 00 P) This is zero when 400 00P 0 because e 0.P 0. Hence P 7. Step To find d π we differentiate dp dπ dp -0.P e (400 00 ) using the product rule. Taking u e 0.P and v 400 00P gives du dv dp dp P 0.P 0.e and 00
Hence 48 d π dv du u + v dp dp dp 0. P 0. P e ( 00) + (400 00 P)( 0.e ) Putting P 7 gives 0.P e (0P 480) d π 00e dp.4 This is negative, so the stationary point is a maximum. 6 To find the price elasticity of demand we need to calculate the values of P, Q and dq/dp. We are given that Q 0 and the demand equation gives P 00 40 ln(0 + ) 78. The demand equation expresses P in terms of Q, so we first evaluate dp/dq and then use the result dq dp d P/dQ To differentiate ln(q + ) by the chain rule we differentiate the outer log function to get Q + and then multiply by the derivative of the inner function, Q +, to get. Hence the derivative of ln(q + ) is and so Q + dp 40 dq Q+ Putting Q 0 gives dp/dq 40/, so that dq/dp /40. Finally, we use the formula
P dq E Q dp 49 to calculate the price elasticity of demand as 78. E.05 0 40 Exercise 4.8 (p. 339) (a) 6e 6x ; (b) 34e 34x ; (c) e x + 4e x ; (d) 40e 4x 4x. () (a) $4885.6; (b) $4887.57; Rate of growth is approximately 4887.57 4885.6 96 0.0 () 0.04t 60 ; e 95.4 3 (a) ; x (b). x 4 (a) 3 x 3x e ; (b) + x + 3x 3 4x 6 x 4 5 (a) 4 + ( 4 + ) x e x e x x e 3 x 4 x 3 4 x (b) lnx + x lnx+ x 6 (a) 4x 4x 4x ( x + ) 4e e x e (x x+ 4) ( x + ) ( x + ) (b) x x (ln x) e e x x e( xlnx ) (ln x) x(ln x)
7 50 Figure S4.5 dy x x x (a) Step e + x ( e ) e ( x) dx At a stationary point x e ( x) 0 x y e Step d y x ( ) ( x e + e ) ( x) e x ( x ) dx When x, d y e < 0 dx Maximum at (, e ); the graph is sketched in Figure S4.5. (b) Step dy dx x At a stationary point, 0 x x x Step dy d y x dx dx x When x, d y 0 dx < Maximum at (, ); the graph is sketched in Figure S4.6.
5 Figure S4.6 8 π TR TC 00ln( Q + ) Q Step Step d π 00 0 Q + 50 Q 49 dq Q + dπ d π 00( Q+ ) 00( + ) dq dq Q When Q 49, d π 00 0 < dq 40 so the point is a maximum. 9 Step dq + dl 0.0 0.0 0.0 700 ( 0.0 L ) 700 L L e e e L (700 4 L) dq At a stationary point, 0 700 4L 0 L 50 dl Step dq 0.0L 0.0L 0.0L 0.0 e (700 4 L) + e ( 4) e (0.8L 8) dl When L 50, dq 4e < 0 dl so the stationary point is a maximum.
0 Q 0 P 00e 5 dp dq e 0.Q 0 so when Q 0, dp dq dq dp 0e 0. e P dq 00e E ( 0.e) Q dp 0 Chapter 5 Partial Differentiation Section 5. Functions of several variables Practice Problems (a) 0; (b) ; (c) ; (d) ; (e) 0; (f). The value of g is independent of the ordering of the variables. Such a function is said to be symmetric. (a) Differentiating 5x 4 with respect to x gives 0x 3 and, since y is held constant, y differentiates to zero. Hence f 0x 0 0x x 3 3 Differentiating 5x 4 with respect to y gives zero because x is held fixed. Also differentiating y with respect to y gives y, so f 0 y y y (b) To differentiate the first term, x y 3, with respect to x we regard it as a constant multiple of x (where the constant is y 3 ), so we get xy 3. The second term obviously gives 0, so
f x 3 xy 0 53 To differentiate the first term, x y 3, with respect to y we regard it as a constant multiple of y 3 (where the constant is x ), so we get 3x y. The second term is a constant and goes to zero, so f 3x y 0 3x y y 3 (a) f xx 60x f f yy yx f 0 xy (b) f xx y 3 f yy 6 x y f f 6xy yx f xy 4 4 f x + 5x x 3 0x x f f f f x x z z 5 y 5, x+,so, at (, 6). x y z z, 4 x y (a) Δx 0., Δy 0.; z ( 0.) + 4(0.) 0.3, so z increases by approximately 0.3. (b) At (, 6), z 4, and at (.9, 6.), z 4.9, so the exact increase is 0.9.
dy y 6 (a) dx x 3y + 54 (b) dy y x y xy 4 d 5 Exercise 5. (p. 360) 34; 75; 0. (a) f x x, f y 0y 4 ; (b) f x 9x, f y e y ; (c) f x y, f y x + 6; (d) f x 6x 5 y, f y x 6 y + 5y. 3 f x 4x 3 y 5 x f y 5x 4 y 4 + y f x (, 0) f y (, ) 7 4 (a) z xy 6x x 4 5 z so at (,0), 6 x Hence Δz 6 0. 0.6 (b) z + y 3 4x y 4 z so at (,0), 4 y Hence Δz 4 ( 0.5) (c) Δz 6 0.+ 4 ( 0.5).6 5 (a) f x 3x +, f y d y ( 3x + ) dx 3x
(b) y x 3 x +, so 55 dy 3x dx Section 5. Partial elasticity and marginal functions Practice Problems Substituting the given values of P, P A and Y into the demand equation gives Q 500 3(0) (30) + 0.0(5000) 430 Q (a) 3, so P 0 E P ( 3) 0.4 430 Q (b), so P A 30 E P A ( ) 0.4 430 Q (c) 0.0, so Y 5000 E Y 0.0 0. 430 By definition, E Y percentage change in percentage change in Q Y so demand rises by 0. 5 0.6%. A rise in income causes a rise in demand, so good is superior.
U x 000 + 5x 4x 56 U 450 + 5x x,so at (38, 500) x U U 948 and 40 x x If working time increases by hour then leisure time decreases by hour, so Δx. Also Δx 5. By the small increments formula ΔU 948( ) + 40(5) 848 The law of diminishing marginal utility holds for both x and x because U x 4< 0 and U x < 0 3 Using the numerical results in Practice Problem, 948 MRCS.06 40 This represents the increase in x required to maintain the current level of utility when x falls by unit. Hence if x falls by units, the increase in x is approximately.06 $4. 4 MP K K and MP L 4L (a) MPL 4L L MRTS MP K K K (b) Q Q K + L K( K) + L(4 L) K L ( K + L ) Q
Exercise 5. (p. 374) 57 Q Y 0.05Y Q When P 5, P A 0 and Y 00 we have Q 555 and 50 Y Hence E Y Y Q 00 50.7 Q Y 555 Q (a) P When P 0, P A 0 and Y 00 we have Q 65 Hence E P P Q 0 0 ( ) Q P 65 65 Q (b) P A Hence E PA PA Q 5 3 ( ) Q P 65 33 A Q (c) 0.Y Y Q so when Y 00 we have 0 Y Hence E Y Y Q 00 40 0 Q Y 65 33 % change in demand 3 3 0.04%; 33 Complementary since E PA is negative so that an increase in the price of the alternative good causes a decrease in demand. 3 Q Y P A Y P Q 0 When P A 0, Y and P 4 we have Q 0 and 0 Y 4
Hence E Y Y Q 0 ; Q Y 0 58 % change in Y / %. 4 U x 3 x x and U x 3 3 x x so at the point (5,8) we have U x 5 and U x 5 (a) 5 37 ΔU + (b) 5 60 /5 MRCS 5/ 5 5 Q Q Q LK + L MP L; MP K + L K L K L MP K 8, MP L 4¼; so when K 7 and L 4 (a) 4 4 57 5 MRTS 8 3 3 (b) This is the value of MRTS so is 5 3. 6 Q MPK 6K + 3L K Q and MPL 6LK L K( MP ) + L( MP ) K(6K + 3L) + L(6LK) 6K3 + 9LK 3(K3 + 3LK) K L Section 5.3 Comparative statics Practice Problems * b+ I C a + b I a S I a a > 0 *
because 0 < a < 59 Hence an increase in I* leads to an increase in C. If a / then C I * Change in C is (a) Substitute C, I, G, X and M into the Y equation to get Y ay + b + I * + G * + X * (my + M * ) Collecting like terms gives ( a + m)y b + I * + G * + X * M * so Y b+ I + G + X M a+ m * * * * Y (b) * X a+ m Y b+ I + G + X M m ( a+ m) * * * * Now a < and m > 0, so a + m > 0. The autonomous export multiplier is positive, so an increase in X * leads to an increase in Y. The marginal propensity to import multiplier is also positive. To see this note from part (a) that Y/ m can be written as Y a+ m and Y > 0 and a + m > 0. (c) Y 0 + 00 + 300 + 50 40 0.8+ 0. 00 Y 0 0.8 0. 3 * X +
and 60 * Δ X 0 so 0 00 Δ Y 0 3 3 3 If d increases by a small amount then the intercept increases and the demand curve shifts upwards slightly. Figure S5. shows that the effect is to increase the equilibrium quantity from Q to Q, confirming that Q/ d > 0. Figure S5. Section 5.4 Unconstrained optimization Practice Problems f x x, f y 6 6y, f xx, f yy 6, f xy 0. Step At a stationary point x 0 6 6y 0 which shows that there is just one stationary point at (0, ).
Step 6 f xx f yy f xy ( 6) 0 < 0 so it is a saddle point. Total revenue from the sale of G is TR PQ (50 Q ) Q 50Q Q Total revenue from the sale of G is TR P Q (95 3 Q ) Q 95Q 3Q Total revenue from the sale of both goods is TR TR + TR 50Q Q + 95Q 3Q Profit is π TR TC Now (50 Q Q + 95Q 3 Q ) ( Q + 3 QQ + Q ) 50Q Q + 95Q 4Q 3QQ π 50 4Q 3 Q, Q π Q 95 8Q 3Q π π 4, 3, Q Q Q π 8 Q Step At a stationary point 50 4Q 3Q 0 95 3Q 8Q 0
that is, 6 4Q + 3Q 50 () 3Q + 8Q 95 () Multiply equation () by 3, and equation () by 4 and subtract to get 3Q 30 so Q 0. Substituting this into either equation () or equation () gives Q 5. Step This is a maximum because and π π 4< 0, 8< 0 Q Q π π π Q Q Q Q ( 4)( 8) ( 3) 3 0 > Corresponding prices are found by substituting Q 5 and Q 0 into the original demand equations to obtain P 45 and P 65. 3 For the domestic market, P 300 Q, so TR PQ 300Q Q For the foreign market, P 00 - Q, so TR P Q 00Q Q Hence TR TR + TR 300Q Q + 00Q Q We are given that
TC 5000 + 00( Q + Q ) 5000 + 00Q + 00Q 63 so Now π TR TC (300Q Q + 00 Q Q ) (5000 + 00Q + 00 Q ) 00Q Q + 00Q Q 5000 π π 00 Q, 00 Q Q Q π π π, 0, Q Q Q Q Step At a stationary point 00 Q 0 00 Q 0 which have solution Q 00, Q 00. Step This is a maximum because and π < 0, Q π < 0 Q π π π Q Q Q Q ( )( ) 0 0 > Substitute Q 00, Q 00, into the demand and profit functions to get P 00, P 50 and π 0000.
64 Exercise 5.4 (p. 40) (a) f f f f f x y xx yy xy 3x 3 3y 3 6x 6y 0 Step At a stationary point, f x x ± x y 0 f y y ± 0 so there are four stationary points at (, ), (, ), (, ) and (, ). Step At (, ), f f 6 > 0 and xx yy f f f 36 > 0 so the point is a minimum. xx yy xy At (, ), f f 6 < 0 and xx yy f f f 36 > 0 so the point is a maximum. xx yy xy At (, ) and (, ), f f f 36 < 0 so these are both saddle points. xx yy xy (b) fx 3x + 3y 6x f 6xy 6y f f f y xx yy xy 6x 6 6x 6 6y Step At a stationary point, f ( x ) y 0 so either x or y 0 y When x the equation f 0 y y ± x When y 0, the equation, fx 0 x x 0 x( x ) 0 x 0 or so there are four stationary points at (, 0), (0, 0), (, ) and (, ). Step At (,0), f f 6 > 0and xx At (0,0), f f 6 < 0 and xx yy yy f f f 36 > 0 so the point is a maximum. xx yy xy f f f 36 > 0 so the point is a minimum. xx yy xy
At (, ) and (, ), At a stationary point, f f f 36 < 0 so these are both saddle points. xx yy xy 65 π 4 Q Q 0 Q+ Q 4 Q π Q Q 4Q + 33 0 Q + 4Q 33 Subtracting twice the second equation from the first gives: 7Q 4 Q 6 Substituting this into the second equation gives Q+ 4 33 Q 9. π π < 0, 4< 0 Q Q π π π > Q Q Q Q ( )( 4) ( ) 7 0 max 3 π TR TC 70Q + 50 Q ( Q + QQ + Q) 70Q + 50Q Q QQ Q π 70 Q Q Q π 50 Q Q Q π Q π Q π Q Q Step At a stationary point Q + Q 70 Q + Q 50 Doubling the second equation and subtracting it from the first gives 3Q 30 Q 0
Substituting this into the second equation gives Q+ 0 50 Q 30 66 Step maximum π π π π π < 0, < 0 3 0 > Q Q Q Q Q Q so the point is a Finally substituting Q 30 and Q 0 into the formula for profit gives $300. 4 U 60 + 5x 0x x U 30 + 5x x x U 0 x U x U x x 5 Step At a stationary point 0x 5x 60 5x + x 30 Adding four times the second equation to the first gives 3x 500 x 500 Substiuting this into the first equation gives 0x 500 60 0x 760 x 38 Step U U U U U 0 < 0; < 0; 3 0 > x x x x x x so the stationary point is a maximum.
The individual works for 30 hours a week and earns $500 so the hourly rate of pay is 67 500 $6.67 30 5 ( ) TR PQ Q Q Q Q Q + Q Q + Q (00 ) 00 00( ) 00Q + 00Q Q Q 4QQ Hence π TR TC 00Q + 00Q Q Q 4 Q Q (8 Q + Q ) 9Q + 00Q Q 3Q 4QQ π 9 4Q 4Q Q π Q 00 6Q 4Q π 4 Q π 6 Q π Q Q 4 Step At a stationary point 4Q + 4Q 9 4Q + 6Q 00 Subtract the second equation from the first: Q 8 Q 4 Subsituting this into the first equation gives 4Q + 6 9 4Q 76 Q 9 Step
π π π π π 4< 0; 6< 0; 8 0 > Q Q Q Q Q Q 68 so the stationary point is a maximum. Section 5.5 Constrained optimization Practice Problems Step We are given that y x, so no rearrangement is necessary. Step Substituting y x into the objective function gives z x xy+ y+ 3 0 z x x + x+ 3 0 x + x+ 0 Step 3 At a stationary point dz 0 dx that is, -x + 0 which has solution x. Differentiating a second time gives d z d x confirming that the stationary point is a maximum.
The value of z can be found by substituting x into 69 z x + x + 0 to get z. Finally, putting x into the constraint y x gives y. The constrained function therefore has a maximum value of at the point (, ). We want to maximize the objective function U x x subject to the budgetary constraint x + 0x 400 Step Step x 00 5x U 00x 5x Step 3 du d x x 00 0 0 has solution x 0. d U dx 0 < 0 so maximum. Putting x 0 into constraint gives x 00. and U U U x 0 x U x 00 x so the ratios of marginal utilities to prices are
U 0 P 0 70 and U 00 P 0 0 which are the same. 3 We want to minimize the objective function TC 3x + xx + 7x subject to the production constraint x + x 40 Step x 40 x Step TC 3(40 x ) + (40 x ) x + 7x 4800 60x + 8x Step 3 d(tc) 60 + 6x 0 dx has solution x 0. d(tc) dx 6 > 0 so minimum. Finally, putting x 0 into constraint gives x 30.
Exercise 5.5 (p. 43) 7 (a) 9x+ 3y 3y 9x (subtract 9x from both sides) y 3 x (divide both sides by 3) 3 (b) z 3xy 3x 3x x 9x 3 At a stationary point, dz 0 8x 0 x dx 9 Hence y and 3 3 3 z 3 9 3 9 d z 8 < 0 so the stationary point is a maximum dx y x y x + z 6x 3x + ( x + ) 6x x + 4 dz At a stationary point, 0 6 x 0 x 3 dx Hence y and z 3. d z < 0 so the stationary point is a maximum dx 3 x + y 500 y 500 x z 80x 0.x + 00(500 x) 0.(500 x) z 80x 0.x + 50000 00x 50000 + 00x 0.x 80x 0.3x dz At a stationary point, 0 80 0.6x 0 x 300 dx Hence y 00 and z 7000. d z 0.6 < 0 so the stationary point is a maximum dx
4 50KL 00 K 4L 7 TC K + 3L 48L + 3L At a stationary point, dtc ( ) + dl 0 48L 3 0 L 6 L 4 Hence K 6 d ( TC) dl 3 L so at L4 the value of 96 d ( TC) 3 > 0 so the stationary point is a dl minimum. 5 x+ y y x TC x + x x x 0 0 8(0 ) 5 (0 ) TC x + x + x x + x x x + 300 30 8 00 5 35 40 300 dtc ( ) At a stationary point 0 70x 40 0 x 6 y 4 dx d ( TC) 70 > 0 min dx 6 x + 4x 360 x 360 4x U x x (360 4 x ) x 360x 4x du At a stationary point, 0 360 8x 0 x 45 U 800 dx du dx 8< 0 so the stationary point is a maximum.
73 Section 5.6 Lagrange multipliers Practice Problems Step g(x, y, λ) x xy + λ ( x y) Step g 4x y λ 0 x g x λ 0 y g x y 0 λ that is, 4x y λ 0 () x λ 0 () x + y (3) Multiply equation () by 4 and add equation (), multiply equation (3) by 4 and subtract from equation () to get y 5λ 0 (4) 5y λ 48 (5) Multiply equation (4) by 5 and subtract equation (5) to get 4λ 48 (6) Equations (6), (5) and () can be solved in turn to get λ, y 0, x
74 so the optimal point has coordinates (, 0). The corresponding value of the objective function is () (0) Maximize U x x + 3x subject to x + x 83 Step g(x, x, λ) x x + 3x + λ(83 x x ) Step g x + 3 λ 0 x g x λ 0 x g 83 x x 0 λ that is, x λ 3 () x λ 0 () x + x 83 (3) The easiest way of solving this system is to use equations () and () to get λ x + 3 and λ x respectively. Hence x x + 3 Substituting this into equation (3) gives 4x + 3 83 which has solution x 0 and so x λ 43.
The corresponding value of U is 75 (43)(0) + 3(43) 849 The value of λ is 43, so when income rises by unit, utility increases by approximately 43 to 89. 3 Step Step g( x, x, λ) x + x + λ( M Px P x ) g x λp 0 x g x λp 0 x g M Px Px 0 λ From equations () and () λ x P and λ x P respectively. Hence that is, x P x P so x P x P x P x (4) P Substituting this into equation (3) gives xp M Px 0 P which rearranges as
x PM P ( P + P ) 76 Substitute this into equation (4) to get x PM P( P + P ) Exercise 5.6 (p. 44) Step g( x, y, λ) x+ xy+ λ(5 x y) Step g + y λ 0 λ y x g x λ 0 λ x y g 5 x y 0 x + y 5 λ From the first two equations we have x y Adding this to the third equation we get x 6 x 3 The constraint gives 3+ y 5 y y Hence the maximum value of z is 9. (a) Step g( x, y, λ) 4 xy+ λ(40 x y) Step g λ 4y λ 0 y x 4 g λ 4x λ 0 x y g 40 x y 0 x + y 40 λ
Substituting the first two equations into the third gives 77 λ+ λ 40 λ 40 and so x 0 and y 0 The minimum value of z is 800 (b) Replacing 40 by 40.5 in part (a) gives λ 4 and so x 0.5, y 0.5 and z 840.5 (c) The change is 40.5 compared to a multiplier of 40 3 Want to maximize Q KL subject to K + L 6 Step g( K, L, λ) KL+ λ(6 K L) Step g L λ 0 L λ K g K λ 0 K λ L g 6 K L 0 K + L 6 λ Substituting the first two equations into the third: 4λ 6 λ.5 and so K.5 and L 3 Hence the maximum level of output is 4.5 4 TR PQ + P Q (50 Q Q ) Q + (00 Q 4 Q ) Q 50Q Q Q Q + 00Q QQ 4Q 50Q + 00Q QQ Q 4Q Hence π TR TC 50Q + 00Q Q Q Q 4 Q (5Q + 0 Q ) 45Q + 90Q Q Q Q 4Q
Step 78 g ( Q, Q, λ) 45Q + 90Q Q Q Q 4 Q + λ(00 5Q 0 Q ) Step g Q g Q 45 Q Q 5λ 0 Q + Q + 5λ 45 90 Q 8Q 0λ 0 Q + 8Q + 0λ 90 g 00 5Q 0Q 0 5Q + 0Q 00 λ The Lagrange multiplier can be eliminated from the first two equations by doubling the first and subtracting the second: Q 4Q 0 Q Q Substituting this into the third equation gives 0Q 00 Q 5 Hence Q 0. Substituting these values into the formula for π shows that the maximum profit is $600. Substituting these values into the first equation gives 30 + 5λ 45 5λ 5 λ 3 Lagrange multiplier is 3, so profit rises to $603 when total cost increases by unit. Chapter 6 Integration Section 6. Indefinite integration Practice Problems (a) x ; (b) x 4 ; (c) x 00 4 ; (d) ; 4 x (e) 9. 9 x
st year $600; nd year $95; 3 rd year $3788; 4 th year $4486 so the first year it 87 happens is the fourth 8 8 8 0.075t 0.075t 000 60000 $790.4 0 0 P e dt e Chapter 7 Matrices Section 7. Basic matrix operations Practice Problems (a), 5, 3 5,. (b), 4, 6,, 6,?, 6; the value of c 43 does not exist, because C has only three rows. 3 4 7 5 A T 0 6 3 5 8 4 0 B T 5 7 9 3 C T 4 5 C 3 5 6 Matrices with the property that C T C are called symmetric. Elements in the top right-hand corner are a mirror image of those in the bottom left-hand corner.
3 (a) 7 ; 3 8 (c) 3 ; (d) ; (e) 0 0. 0 0 88 Part (b) is impossible because A and C have different orders. 4 () (a) 4 6 0 ; 0 8 (b) 4 4 ; (c) 3 5 ; 0 (d) 6 0 4. 0 From (a) and (b) 4 0 6 A + B 6 0 4 4 0 4 + 0 8 0 which is the same as (d), so (A + B) A + B () (a) 3 6 9 5 ; 0 (b) 6 8 30. 0 4 From (a), 3 6 6 (3A) 9 5 8 30 0 0 4 which is the same as (b), so (3A) 6A 5 (a) [8] because (0) + ( )( ) + 0() + 3() + () 8 (b) [0] because ( ) + () + 9(0) 0.
(c) This is impossible, because a and d have different numbers of elements. 89 Figure S7. 6 AB c c 0 c c 3 4 3 c3 c 3 AB 7 c 0 c c 3 4 3 c3 c 3 7 0 AB 0 c c 3 4 3 c3 c 3 7 0 AB 0 3 c 3 4 3 c3 c 3 AB 7 0 0 3 4 3 4 3 c c 3 3 7 0 AB 0 3 4 3 4 3 6 c 3
AB 7 0 0 3 4 3 4 3 6 0 90 7 (a) 5 7; 5 (d) 4 3 ; 5 5 (f) 9 6 3 ; 7 5 8 (g) 5 7 9 3 3 3 ; 6 9 (h) 5 6. 5 Parts (b), (c) and (e) are impossible because, in each case, the number of columns in the first matrix is not equal to the number of rows in the second. 8 Ax is the 3 matrix x+ 4y+ 7z z 6y 5z + + 8x+ 9y+ 5z However, x + 4y + 7z 3, x + 6y + 5z 0 and 8x + 9y + 5z, so this matrix is just 3 0 which is b. Hence Ax b. Exercise 7. (p. 478) (a) J 35 7 3 ; 4 39 4 F 3 7 3. 5 9 6 (b) 66 44 6 67 68 40
(c) 4 0 0 7 0 8 9 4 6 8 (a) 0 0 0 4 6 8 (b) (c) 4 8 4 0 0 8 0 6 4 0 0 30 6 0 0 4 6 4 (d) Same answer as (c). 3 4B, (CB) T, CBA are possible with order 3, 3 4, 4 respectively. 4 (a) 5900 00 Total cost charged to each customer. (b) 3 7 3 3 4 5 Amount of raw materials used to manufacture each customer s goods. (c) 35 75 30 Total raw material costs to manufacture one item of each good. (d) 005 05 Total raw material costs to manufacture requisite number of goods for each customer.
(e) [7000] 9 Total revenue received from customers. (f) [0] Total cost of raw materials. (g) [5790] Profit before deduction of labour, capital and overheads. 5 () (a) 3 5 4 6 3 (b) 4 (c) (d) 5 5 0 5 5 0 (A + B) T A T + B T : that is, transpose of the sum is the sum of the transposes. () (a) (b) (c) (d) 5 4 9 0 0 4 5 9 5 4 9
(CD) T D T C T : that is transpose of a product is the product of the transposes multiplied in 93 reverse order. 6 (a) B + C 0 6 5 so A(B + C) 5 4 5 4 AB AC 7 5 6 0 8 4, so and AB + AC 5 4 5 4 (b) AB (AB)C 7 5 6 0, so 3 43 4 6 BC 4 4 4, so A(BC) 3 43 4 6 7 AB [ 3]; BA 4 0 7 4 8 3 6 9 4 8 6 8 (a) 7x + 5y x + 3y
(b) A 3, 4 5 x x y, b z 6 3. 94 Section 7. Matrix inversion Practice Problems A 6() 4() 8 0 so A is non-singular and its inverse is given by 4 4 8 6 8 3 4 B 6() 4(3) 0 so B is singular and its inverse does not exist. We need to solve Ax b, where 9 P 43 A 7 x P b 57 Now A - 7 6 9 so P 7 43 4 P 6 9 57 7 3 In equilibrium, Q S Q D Q, say, so the supply equation becomes P aq + b Subtracting aq from both sides gives P aq b () Similarly, the demand equation leads to
95 P + cq d () In matrix notation equations () and () become a P b c Q d The coefficient matrix has an inverse, c a c+ a so that P c a b Q c+ a d that is, cb + ad P c+ a and b+ d Q c+ a The multiplier for Q due to changes in b is given by the element in row, column, of the inverse matrix so is c+ a Given that c and a are both positive it follows that the multiplier is negative. Consequently, an increase in b leads to a decrease in Q.
4 4 3 A + 7 3 4 96 A A A A A A A A 3 3 3 3 33 3 4 4 + 3 3 3 3 4 3 3 + 4 3 0 3 3 3 + 3 4 3 3 3 0 3 + 4 5 Expanding along the top row of A gives A a A + a A + a A 3 3 (7) + 3( ) + 3( ) using the values of A, A and A 3 from Practice Problem 4. Other rows and columns are treated similarly. Expanding down the last column of B gives B b B + b B + b B 3 3 3 3 33 33 0( B ) + 0( B ) + 0( B ) 0 3 3 33 6 The cofactors of A have already been found in Practice Problem 4. Stacking them in their natural positions gives the adjugate matrix 7 3 0 3 0
Transposing gives the adjoint matrix 97 7 3 3 0 0 The determinant of A has already been found in Practice Problem 5 to be, so the inverse matrix is the same as the adjoint matrix. The determinant of B has already been found in Practice Problem 5 to be 0, so B is singular and does not have an inverse. 7 Using the inverse matrix in Practice Problem 6, P 7 3 3 3 8 P 0 37 5 P 3 0 35 3 Exercise 7. (p. 497) () (a) A 3; (b) B 4; (c) 4 4 AB 7 4 so AB. These results give AB A B : that is, determinant of a product is the product of the determinants. () (a) (b) (c) /3 /3 A 5/3 /3 0 B / /4 /3 /3 ( AB ) 7/ /3
These results give (AB) B A : that is, inverse of a product is the product of the 98 inverses multiplied in reverse order. a + 3 0 a 3/ 8 3b 0 b 8/3 3 (a) (b) x 4 y 3 5 3 6 x 3 8 y 3 4 6 Hence x, y ; Hence x, y. 4 9 ; 5 3 P 9 0 40 ; P 5 3 0 0 P 40; P 0 5 (a) 50 P+ P 0 + P 3P P 70 0 + P 4P 0 + 5P P + 9P 0 (b) Inverse 9 6 3 9 70 5 6 3 0 5 P 5, P 5
Section 7.3 Cramer s rule 99 Practice Problems (a) By Cramer s rule det(a ) x det(a) where 6 det( A) 66 3 9 4 det( A) 3 5 x Hence 66 3 (b) By Cramer s rule det( A3) x 3 det( A) where 3 4 8 det( A ) 5 4 3 9 5 4 4 5 4 + 8 9 3 9 3 4(37) ( 30) + 8( 9) 6 and
4 3 det( A) 5 3 4 5 5 4 + 3 4 3 4 3 4(8) ( ) + 3( 9) 6 00 Hence 6 x 3 6. The variable Y d is the third, so Cramer s rule gives det( A3) Y d det( A) where A 3 * * I + G 0 0 b 0 0 0 * t 0 T Expanding along the second row gives 3 I + G * * 0 0 det( A ) 0 b 0 0 0 * t T t since along the second row the pattern is + +. Now * * 0 I + G 0 4 * * 0 ( I + G ) * * T t t T * * * T ( I + G )( + t) (expanding along the first row) and
0 0 ( ) + t t t 0 0 (expanding down the second column). Hence det(a 3 ) T* (I* + G*)( + t) b( + t) From the worked example given in the text, det(a) a + at Hence y d * * * T ( I + G )( + t) b( + t) a+ at 3 Substituting C, M and I* into the equation for Y gives Y 0.7Y + 50 + 00 + X 0.3Y Also, since X M 0.Y, we get Y 0.7Y + 50 + 00 + 0.Y 0.3Y which rearranges as 0.6Y 0.Y 50 In the same way, the second set of equations leads to 0.3Y + 0.3Y 400 Hence 0.6 0. Y 50 0.3 0.6 Y 400 In this question both Y and Y are required, so it is easier to solve using matrix inverses rather than Cramer s rule, which gives
0.3 0. 50 Y 0.5 0. 0.6 400 5 0.5 35 0 Hence Y 766.67 and Y 00. The balance of payments for country is X M M M 0.Y 0.3Y 0.(00) 0.3(766.67) 0 Moreover, since only two countries are involved, it follows that country will have a surplus of 0 Exercise 7.3 (p. 508) 4 3 (a) 7 5 4 x 7 3 4 5 (b) (c) 5 4 5 3 x ; 3 4 3 5 9 4 3 7 75 x 5 4 5 7 (a) 9 0 y 3 0 4
(b) 5 7 9 y 5 9 3 03 (c) 7 3 9 y 3 9 3 5 3 (a) 3 4 3 5 4 3 4 x ; y 4 3 4 4 3 4 5 5 (b) (c) 3 4 5 8 4 x ; y 3 4 3 4 4 3 4 5 5 3 4 36 5 98 36 40 x 7; y 0 4 3 4 4 3 4 5 5 4 (a) 400 5P 3P 60 + 3P 8P + 3P 460 300 P 3P 00 + P P + 5P 400 (b) 460 3 400 5 00 550 6 3 8 3 34 7 7 5 5 (a) Y C I* ay + C b
* Y I a C b 04 (b) I a b b+ ai C a * a * Chapter 8 Linear Programming Section 8. Graphical solution of linear programming problems Practice Problems The line x + 3y 6 passes through (0, ) and ( 6, 0). Substituting x, y 4 into the equation gives + 3(4) This is greater than 6, so the test point satisfies the inequality. The corresponding region is shown in Figure S8.. Figure S8.