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SOLUTION We let x, y, and z to be the length, width, and height, respectively, of the box in meters. Then we wish to maximize V = xyz subject to the constraint g(x, y, z) = 2xz + 2yz + xy = 48. Using the method of Lagrange multipliers, we look for the values of x, y, z, and A such that VV = AVg and g(x, y, z) = 48. This gives the equations Vy = Ag, Vy = Agy Vz = Ag, 2xz + 2yz + xy = 48 which become (1) yz = a (2) = a(2z + x) (3) xy = A (4) 2xz + 2yz + xy = 48. There are no general rules for solving systems of equations. Sometimes some ingenuity is required. In the present example, you might notice that if we multiply (1) by x, (2) by y, and (3) by z, then the left sides of these equations will be identical. Doing this, we have (5) xyz = A(2xz + xy) (6) xyz = A(2yz + xy) (7) xyz = A(2xz + 2yz). We observe that A * 0 because A = 0 would imply that yz = xz = xy = 0 from (1), (2), and (3) and this would contradict (4). Therefore, from (5) and (6), we have 2xz + xy = 2yz + xy which gives xz = But z + 0 (since z = 0 would give V = 0), so x = From (6) and (7) we have 2yz + xy = 2xz + 2yz which gives 2xz = . and so (since x * 0) y = If we now put x = y = 2z in (4) we get 4z2 + 4z2 + 4z2 = 48. Since x, y and z are all positive, we therefore have z = and so x = and y =

SOLUTION We let x, y, and z to be the length, width, and height, respectively, of the box in meters. Then we wish to maximize V = xyz subject to the constraint g(x, y, z) = 2xz + 2yz + xy = 48. Using the method of Lagrange multipliers, we look for the values of x, y, z, and A such that VV = AVg and g(x, y, z) = 48. This gives the equations Vy = Ag, Vy = Agy Vz = Ag, 2xz + 2yz + xy = 48 which become (1) yz = a (2) = a(2z + x) (3) xy = A (4) 2xz + 2yz + xy = 48. There are no general rules for solving systems of equations. Sometimes some ingenuity is required. In the present example, you might notice that if we multiply (1) by x, (2) by y, and (3) by z, then the left sides of these equations will be identical. Doing this, we have (5) xyz = A(2xz + xy) (6) xyz = A(2yz + xy) (7) xyz = A(2xz + 2yz). We observe that A * 0 because A = 0 would imply that yz = xz = xy = 0 from (1), (2), and (3) and this would contradict (4). Therefore, from (5) and (6), we have 2xz + xy = 2yz + xy which gives xz = But z + 0 (since z = 0 would give V = 0), so x = From (6) and (7) we have 2yz + xy = 2xz + 2yz which gives 2xz = . and so (since x * 0) y = If we now put x = y = 2z in (4) we get 4z2 + 4z2 + 4z2 = 48. Since x, y and z are all positive, we therefore have z = and so x = and y =

College Algebra (MindTap Course List)
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter6: Linear Systems
Section6.8: Linear Programming
Problem 36E
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SOLUTION We let x, y, and z to be the length, width, and height, respectively, of the box in meters. Then we wish to maximize V = xyz subject to the constraint g(x, y, z) = 2xz + 2yz + xy = 48. Using the method of Lagrange multipliers, we look for the values of x, y, z, and A such that VV = AVg and g(x, y, z) = 48. This gives the equations Vy = Ag, Vy = Agy Vz = Ag, 2xz + 2yz + xy = 48 which become (1) yz = a (2) = a(2z + x) (3) xy = A (4) 2xz + 2yz + xy = 48. There are no general rules for solving systems of equations. Sometimes some ingenuity is required. In the present example, you might notice that if we multiply (1) by x, (2) by y, and (3) by z, then the left sides of these equations will be identical. Doing this, we have (5) xyz = A(2xz + xy) (6) xyz = A(2yz + xy) (7) xyz = A(2xz + 2yz). We observe that A * 0 because A = 0 would imply that yz = xz = xy = 0 from (1), (2), and (3) and this would contradict (4). Therefore, from (5) and (6), we have 2xz + xy = 2yz + xy which gives xz = But z + 0 (since z = 0 would give V = 0), so x = From (6) and (7) we have 2yz + xy = 2xz + 2yz which gives 2xz = . and so (since x * 0) y = If we now put x = y = 2z in (4) we get 4z2 + 4z2 + 4z2 = 48. Since x, y and z are all positive, we therefore have z = and so x = and y =
Transcribed Image Text:SOLUTION We let x, y, and z to be the length, width, and height, respectively, of the box in meters. Then we wish to maximize V = xyz subject to the constraint g(x, y, z) = 2xz + 2yz + xy = 48. Using the method of Lagrange multipliers, we look for the values of x, y, z, and A such that VV = AVg and g(x, y, z) = 48. This gives the equations Vy = Ag, Vy = Agy Vz = Ag, 2xz + 2yz + xy = 48 which become (1) yz = a (2) = a(2z + x) (3) xy = A (4) 2xz + 2yz + xy = 48. There are no general rules for solving systems of equations. Sometimes some ingenuity is required. In the present example, you might notice that if we multiply (1) by x, (2) by y, and (3) by z, then the left sides of these equations will be identical. Doing this, we have (5) xyz = A(2xz + xy) (6) xyz = A(2yz + xy) (7) xyz = A(2xz + 2yz). We observe that A * 0 because A = 0 would imply that yz = xz = xy = 0 from (1), (2), and (3) and this would contradict (4). Therefore, from (5) and (6), we have 2xz + xy = 2yz + xy which gives xz = But z + 0 (since z = 0 would give V = 0), so x = From (6) and (7) we have 2yz + xy = 2xz + 2yz which gives 2xz = . and so (since x * 0) y = If we now put x = y = 2z in (4) we get 4z2 + 4z2 + 4z2 = 48. Since x, y and z are all positive, we therefore have z = and so x = and y =
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