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sola7

Course: MATH 1052, Fall 2009
School: Allan Hancock College
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Word Count: 1155

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Semester MATH1052 1 - 2009 Solutions to Assignment 7 1. a) Let g(x, y) = x2 + 2y 2 so the constraint is g(x, y) = 44. Now we have f = g 3i 2j = (2xi + 4yj). 2x 3 x = or y = . Now using 4y 2 3 So 2x = 3 and 4y = 2. This leads to 2 2x2 the constraint we see that x + = 44. So we obtain 9 11x2 = 44 9 x2 = 36 x = 6 and y= 2. So f (6, 2) = 18 + 4 = 22 is maximum and f (6, 2) = 18 4 = 22 is minimum. b) Let...

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Semester MATH1052 1 - 2009 Solutions to Assignment 7 1. a) Let g(x, y) = x2 + 2y 2 so the constraint is g(x, y) = 44. Now we have f = g 3i 2j = (2xi + 4yj). 2x 3 x = or y = . Now using 4y 2 3 So 2x = 3 and 4y = 2. This leads to 2 2x2 the constraint we see that x + = 44. So we obtain 9 11x2 = 44 9 x2 = 36 x = 6 and y= 2. So f (6, 2) = 18 + 4 = 22 is maximum and f (6, 2) = 18 4 = 22 is minimum. b) Let g(x, y, z) = x2 + y 2 + z 2 . Then f = g 2xi 2j + 4zk = (2xi + 2yj + 2zk) 2 = 2y, and 4z = 2z. 2x = 2x, x = z = 0 and f (0, 1, 0) = 2. 3 3 5 Or = 2, y = 1/2, x = 0, z = and f (0, 1/2, )= . 2 2 2 3 5 So fmax (0, 1/2, )= . 2 2 c) Let g(x, y) = x + y. Then f = g 2x 8 = , (2x 8)i + (2y 12)j = (i + j) y = x + 2. So either = 1, y = 1, 2y 12 = = 2x 8 Now using the constraint x + y = 8 we obtain x + x + 2 = 8 or x = 3 and y = 5. So fmin (3, 5) = 9 24 + 25 60 + 48 = 2. d) Note that this problem has two constraints. As per Stewart, the extreme values will be found when f = g+ h where , are the Lagrange multipliers and g, h are the constraint functions. Component by component we get: fx = gx + hx , fy = gy + hy , fz = gz + hz , g(x, y, z) = 1, h(x, y, z) = 4. These equations give, respectively, 1 = 1 + 0, 2 = 1 + 2y 0 = 1 + 2z x+y+z =4 y2 + z2 = 4 1 The rst equation gives = 1. Substitute into the next two equations to give 2y = 1 and 2z = 1, and dividing one by other, y = z. Substitute this the 2 into the nal equation and 2z = 4, so z = 2, and consequently y = 2 and x = 1. So we have the extreme values at (1, 2, 2) and (1, 2, 2). Look at the function values to nd that the former is a max and the latter is a min. 2. First we note that P = 100(0.25)x0.75 y 0.75 i + 100(0.75)x0.25 y 0.25 j Now P = g implies that y 25( )0.75 = 48, x So y =4 x and x 75( )0.25 = 36 y 25 y 0.75 75 x 0.25 ( ) = ( ) . 48 x 36 y and g = 48i + 36j. y = 4x. Now applying the constraint leads to g(x, 4x) = 48x + 36 4x = 100, 000 192x = 100, 000. So x= Therefore Pmax = 100( 100, 000 0.75 100, 000 0.25 400, 000 0.75 ) ( ) = 100( )4 = \$147313.9128. 192 192 192 100, 000 192 and y= 400, 000 . 192 3. The horizontal velocity is initially u0 = 500 cos /6 and vertically v0 = 500 sin /6. The equation of motion horizontally is mx = 0 and vertically is my = mg. These equations have the following solutions, respectively. x = u0 and y = gt + v0 . . .. .. x = u0 t + x 0 y= 1 2 gt + v0 t + y0 . 2 1 2 gt + v0 t. Let x0 = 0 and y0 = 0 then x = u0 t and y = 2 1 2 gt + v0 t = 0. This implies The rocket hits the ground when y = 0, ie when 2 2v0 2v0 2u0 v0 t = 0 or t = . If t = then x = is the distance from launch pad. g g g v2 v0 is y = 0 . The maximum height when t = g 2g Now put in u0 = 500 cos /6 and v0 = 500 sin /6. We see that the Rocket hits 1000 sin /6 = 51.02 seconds. Distance from launch pad the ground when t = 9.8 2(500)2 sin /6 cos /6 (500)2 sin2 /6 is = 22, 092 meters. Maximum height is = 9.8 2 9.8 3188.8 meters. In our model, we have assumed constant gravity of 9.8 m/s2 and no air resistance and perfectly at ground. 2 4. Same set up as Q3 but now u0 = 20 cos /4 = 2 10 and v0 = 20 sin /4 = 10 2. 1 2 x Then x = 10 2t and y = gt + 10 2t. From rst equation we nd t = 2 10 2 gx2 and now by second equation we obtain y = + x. This is the curve traced out 400 in xy-plane. 5. a) Equilibrium solutions are y(1 y) = 0 y = 0 or 1. y y y y y y =0 = 0.5 =1 = 1.5 = 0.5 the the the the the slope = 0 slope = 1/2 slope = 0 slope = 3/2 slop = 3/2 y=1 t Note that y = 0 is unstable as nearby solutions move away but y = 1 is stable. b) Equilibrium solutions are constant solutions that satisfy t + y = 0 y = t which is not constant. So there is no equilibrium solutions. y y y y y = t = t + 1 = t 1 = t + 2 the the the the slope slope slope slope =0 =1 = 1 =2 t c) Equilibrium solution is y = 4. 3 y y=4 y y y y =1 =2 =4 =5 the the the the slope slope slope slope =3 =2 =0 = 1 t y = 4 is a stable equilibrium solution. d) Equilibrium solutions are y = 1, 2 and 3. y y=3 y=2 y=0 1 y= 2 y=1 3 y= 2 5 y= 2 7 y= 2 y = 2 the slope the slope the slope the slope the slope the slope the slope =3 45 = 16 =2 15 = 16 7 = 16 27 = 16 = 10 t y=-1 6. Stewart 6th Ed - Section 10.2, Q3-6. Q3 y = 2 y. First note that y = 0 when y = 2, but both (I) and (III) have this property. However y only depends on y not x, and moreover if y > 2 then y < 0, and only (III) satises either of these properties. Q4 y = x(2 y). Here y = 0 when y = 2 and when x = 0. This is graph (I). Q5 y = x + y 1. Here y = 0 when x + y = 1, and y > 0 above this line and y < 0 below this line. This is graph (IV). Q6 y = sin x sin y. At x, y = 0, we have y = 0, which only occurs in (II). Stewart 5th Ed - Section 10.2, Q3-6. Q3) y = y 1. Note that at y = 1, we have y = 0, and this is only demonstrated in (IV). Q4) y = y x. Note that along y = x we have y = 0 and this is the case in (I),(II) and (III). When y > x, we have y > 0 and when y &...

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