text: ### Math 484.2     October 2, 2008     Name: ______Patrick Johnson_____corr by L.V and _E.L.__________ Midterm 1, 5 problems, 15 points each. Return this page.
  
  1. Solve for x,y:
  
   x y z 1 2 0 = u 1 0 2 = v
  
  Solution — Pivoting
  
   x y z 1 2 0 = u 1* 0 2 = v
  
  -->
  
   v y z 1 2* -2 = u 1 0 -2 = x
  
  -->
  
   v u z -1/2 1/2 1 = y 1 0 -2 = x
  
  x = v – 2z
  y = -v/2 + u/2 + z
  
  2. 4x + y^2 –> min
  x^2 + y^2 = 1
  
  Solution
  solve eq. 2 for y^2:
  y^2= 1 – x^2
  substitute into eq. 1:
  4x + (1 – x^2) –> min
  -x^2 + 4x + 1 –> min
  Objective funcion is now defined only by x
  Eq. 2 is the formula for a circle of radius 1, so our feasible range of x is -1 <= x <= 1.
  The objective function is minimal when x = -1, with an optimal solution
  value of -4. Then y = 0.
  
  3. Solve for x,y
  
   x y 1 1 1 -x >= 3 -1 1 2 <= 22 1 2 0 –> min
  
  Solution: write equation from row 1:
  x + y – x >= 3
  y >= 3
  From row 2:
  -x + y + 2 <= 2
  -x +y <= 0
  y <= x
  x >= y
  Combining the equations, x >= y >= 3
  Objective function is 2x + y, so graphical method can be used or
  it should be obvious that minimal values of x and y will result in
  minimal objective function, so:
  min = 9 when x = y = 3
  
  4. Solve the linear programs given by the following standard tableaux:
  
  4A:
  
   a b c 1 1 0 -1 1 = d 1 0 1 -2 –> min
  
  Tableau is Optimal
  min = -1(in fact, -2)
  when a,b,c = 0, and d = 1
  
  4B:
  
   a b c 1 -2 -1 -1 -3 = d 1 0 1 -1 –> min
  
  Tableaux is infeasible
  min = + infinity
  
  4C:
  
   a b c 1 -1 0 1 0 = d -2 1 -1 -1 –> min
  
  Tableau (The LP)   is unbounded
  min = – infinity
  
  5. Find all logical implications between the following 5 constraints on x, y:
  
  (a) x_ = y_
  (b) 0 > 1
  (c) 2y > x
  (d) 1/3 > 1/2
  (e) y >= x
  
  Solution:
  (b) is unconditionally true, so (b) is implied by anything
  (d) is unconditionally false, so (d) implies everything
  a => b
  c => b
  d => a
  d => b
  d => c
  d => d
  d => e
  e => b
  
  a => e  (added by L.V.)
explanation: The document is a midterm exam for Math 484.2, dated October 2, 2008, and completed by Patrick Johnson. It contains five problems, each with a question and a corresponding solution.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: ### Math 484.2     October 2, 2008     Name: ______Patrick Johnson_____corr by L.V and _E.L.__________ Midterm 1, 5 problems, 15 points each. Return this page.
  
  1. Solve for x,y:
  
   x y z 1 2 0 = u 1 0 2 = v
  
  Solution — Pivoting
  
   x y z 1 2 0 = u 1* 0 2 = v
  
  -->
  
   v y z 1 2* -2 = u 1 0 -2 = x
  
  -->
  
   v u z -1/2 1/2 1 = y 1 0 -2 = x
  
  x = v – 2z
  y = -v/2 + u/2 + z
  
  2. 4x + y^2 –> min
  x^2 + y^2 = 1
  
  Solution
  solve eq. 2 for y^2:
  y^2= 1 – x^2
  substitute into eq. 1:
  4x + (1 – x^2) –> min
  -x^2 + 4x + 1 –> min
  Objective funcion is now defined only by x
  Eq. 2 is the formula for a circle of radius 1, so our feasible range of x is -1 <= x <= 1.
  The objective function is minimal when x = -1, with an optimal solution
  value of -4. Then y = 0.
  
  3. Solve for x,y
  
   x y 1 1 1 -x >= 3 -1 1 2 <= 22 1 2 0 –> min
  
  Solution: write equation from row 1:
  x + y – x >= 3
  y >= 3
  From row 2:
  -x + y + 2 <= 2
  -x +y <= 0
  y <= x
  x >= y
  Combining the equations, x >= y >= 3
  Objective function is 2x + y, so graphical method can be used or
  it should be obvious that minimal values of x and y will result in
  minimal objective function, so:
  min = 9 when x = y = 3
  
  4. Solve the linear programs given by the following standard tableaux:
  
  4A:
  
   a b c 1 1 0 -1 1 = d 1 0 1 -2 –> min
  
  Tableau is Optimal
  min = -1(in fact, -2)
  when a,b,c = 0, and d = 1
  
  4B:
  
   a b c 1 -2 -1 -1 -3 = d 1 0 1 -1 –> min
  
  Tableaux is infeasible
  min = + infinity
  
  4C:
  
   a b c 1 -1 0 1 0 = d -2 1 -1 -1 –> min
  
  Tableau (The LP)   is unbounded
  min = – infinity
  
  5. Find all logical implications between the following 5 constraints on x, y:
  
  (a) x_ = y_
  (b) 0 > 1
  (c) 2y > x
  (d) 1/3 > 1/2
  (e) y >= x
  
  Solution:
  (b) is unconditionally true, so (b) is implied by anything
  (d) is unconditionally false, so (d) implies everything
  a => b
  c => b
  d => a
  d => b
  d => c
  d => d
  d => e
  e => b
  
  a => e  (added by L.V.)
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a midterm exam for Math 484.2, dated October 2, 2008, and completed by Patrick Johnson. It contains five problems, each with a question and a corresponding solution.
  
  Label: multiple-question-and-answer