x+ y = 50 Dividing both sides by 2 : ( ) dx By (7.2), x = 25m gives maximum area. Substituting this value into (*):
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1 Solutions 7(b 1 Complete solutions to Exercise 7(b 1. Since the perimeter 100 we have x+ y 100 [ ] ( x+ y 50 ividing both sides by : y 50 x * The area A xy, substituting y 50 x gives: A x( 50 x A 50x x ifferentiating to find the stationary point: 50 x 0 gives x 5m dx d A < 0 dx By (7., x 5m gives maximum area. Substituting this value into (*: y m It seems as if the area constraint by the fence needs to be a square, x y 5m.. From the fencing of 40 m around the field we have x + y 40 y 40 x ( The area A is given by A xy x( 40 x by ( 40 x x For stationary points: 40 4x 0, 4x 40, x 60m dx To show maximum we need to differentiate again d A dx 4 < 0 By (7., when x 60m we have maximum area. To find y we substitute x 60 into ( ; y 40 ( 60 10m Observe that y is twice the length of x, x 60m and y 10m gives maximum area.. We can rewrite + k as + k C C C 1 + k ifferentiating this with respect to gives: d ( C dc C 1 + k C + k k (7. A 0, A < 0 For stationary point this is equal to zero: maximum
2 Solutions 7(b k 0 gives k Rearranging gives C. Taking the square root of both sides: k k C 1 k To check that this value of gives a minimum we need to differentiate again: d ( k C dc ( C ( Substituting C 1 k into ( gives: 1 ( C k ( C k ( k C / k C / k C C C C C C k k 1 k k > 0 ( taking the positive square root Hence by (7. when k we have minimum drag. 4. We have πr h 8 h 8 πr ( The surface area, A, is given by the base, πr, and the curved area, πrh, hence A πr + πrh πr 8 + πr πr πr + 16 r For stationary points: dr πr 16r 0 πr 16 r A πr +16r 1 r 16 π 8 π 8 Take the cube root of both sides: r 8 π π π 1 To find whether this stationary point is a maximum or minimum we use the second derivative test: d A πr 16r π + r dr dr (7. A 0, A > 0 minimum
3 Solutions 7(b Substituting r π into d A 1 dr gives d A > 0. By (7., when r dr π 1m we have a minimum surface area. How do we find h? Substitute r 1 into : π ( 8 h 1 π π ( 8 π π ( π The height and radius are equal. 5. Similar to EXAMPE 9 with 1000 replaced by V. Volume V πr h, gives V h (* π r Surface Area A πr + πrh πr V + πr πr A πr + Vr 1 4π r Vr 0 dr V 4π r r 1 1 V V V V, r r 4π π π π To show we have minimum surface area we have to differentiate again d A 4π + 4Vr dr 4V 4π + > 0 r d A dr is going to be positive because r is radius and is therefore positive. By (7., when r V 1 π we have minimum surface area. To find h we substitute r V 1 into (* and obtain π h r. 6. Factorizing the given equation gives: W M ( x x dm W ( x 0, so x 0 which gives x dx dm W ( W < 0 dx (7. M 0, M < 0 (7. A 0, A < 0 maximum minimum
4 Solutions 7(b 4 By (7., the bending moment is a maximum at x. The maximum value of M is evaluated by substituting x into M : W M ( x x W. 7. We have W W W y W ( 6EI x x ( * For stationary points dy W ( x 0, x 0, x, x ± dx 6EI x cannot be because is a length. Hence x. For maximum; d y W ( 6x dx 6EI d y W x 0 < dx EI By (7., at x we have maximum deflection. Maximum deflection can be evaluated by substituting x into (* W y ( 6EI W ( 6EI W W ( 6EI EI 8. Clearly by looking at Fig 0 it looks as if the maximum deflection will occur furthest from the fixed end, x. We need to show this by using differentiation. W 4 y x 4x 4 x 4EI + dy W 4Wx 4x 1x 8 x x x dx 4EI + + 4EI 0 factorizing a factor of 4x x 0 or x x+ 0 ( x ( x 0 x, x We have a stationary points at x 0, x and x. x cannot be because the cantilever is only of length. Also at x 0, the fixed end, there is no deflection (minimum. We need to show that at x we do have maximum deflection. (7. y 0, y < 0 maximum
5 Solutions 7(b 5 dy W 4x 1x 8 x dx 4EI + d y W 1x 4x + 8 dx 4EI Substituting x gives d y W [ dx 4EI ] W [ 4EI 4 ]< 0 By (7., at x we have maximum deflection. The value of this deflection is W 4 4 y 4 4 4EI + 4 W 4 W 4EI 4EI 9. We have p Tv mv dp T mv 0, mv T dv transposing gives T v m To show maximum for this value of v : d p T 6mv 6m dv m < 0 T Hence by (7., v gives maximum power. m 10. (i T sin( θ + 6sin( θ dt 6cos( θ + 6cos( θ dθ ( ( θ ( θ 6 cos 1 + 6cos by (4.54 ( θ + ( θ We need to solve 1 cos ( θ + 6 cos( θ 6 0 et x cos( θ : 1x + 6x 6 0 1cos 6cos 6 [ ] x + x 1 0 ividing by 6 ( x 1 ( x +1 0 gives x 1 or x 1 If cos( θ 1 then θ π. If cos( θ 1 then θ lies outside the range 0 θ < π. So θ can only be π. dt 1sin ( θ 6sin ( θ dθ d T At θ π, dθ 1sin π 6sin π < 0 By (7., at θ π we have maximum torque. (4.54 ( x ( cos cos x 1 (7. y 0, y < 0 maximum
6 Solutions 7(b 6 π π 9 (ii T sin + 6sin 7.8Nm 11. (i Similar to EXAMPE 10. ( P i R + i i R For stationary points: dp ir 1 1+ ( i i1 R( 1 0 di 1 ir i1 (* R1+ R To establish this stationary point is a minimum, use the nd derivative: dp R1+ R > 0 ( because R1 > 0 and R > 0 di1 By (7. (* gives min P. (ii Substitute for i 1 from (* into ir i 1 R 1 gives the required result.
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