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Problem-4 (Multi-Source-Integration between Curves)

Curve Set I

Curve Set II

Curve Set III

Curve 1:
Z = 2x + 2y

 

Curve 2:
Z = 4x - 3y

Curve 3:
Z = 2y - x

 

Curve 4:
Z = 2y + 3x


Curve 5:
Z = 2x - 3y

Curves A, B, C, and, D

 

A = Maximum of the values of Z from Curve Set I.

 

B = Maximum of the values of Z from Curve Set II.

 

C = Minimum of the values of Z from Curve Set I.

 

D = Minimum of the values of Z from Curve Set II.

 

E = Maximum of the values of A, B, C, and D above + the Minimum of the values of A, B, C, and D above.

 

F = Maximum of the values of A, B, C, and D above the Minimum of the values of A, B, C, and D above.

  1. If x = –3 and y = 4, then identify the expressions whose values are maximum or minimum?
Maximum Minimum  
A
B
C
D
E
  1. In the table below, choose the correct implications of A being positive, and C being negative, separately.
C is negative E is greater than F  
A is positive.
B is positive.
D is positive.
E is greater than A, B, C, D, and F.
F is greater than A, B, C, D, and E.

Solution-4 (Multi-Source-Integration between Curves)

  1. Substituting x = –3 and y = 4 yields

Curve Set I

Curve Set II

Curve Set III

Curve 1:

Z = 2x + 3y
  = 2(–3) + 3(4)
  = –6 + 12
  = 6

 

Curve 2:

Z = 4x – 3y
  = 4(–3) – 3(4)
  = –12 – 12
  = –24

Curve 3:

Z = 2y x
  = 2(4) – (–3)
  = 8 + 3
  = 11

 

Curve 4:

Z = 2y + 3x
  = 2(4) + 3(–3)
  = 8 – 9
  = –1

 

Curve 5:

Z = 2x – 3y
  = 2(–3) – 3(4)
  = –6 – 12
  = –18

Curves A, B, C, and, D

 

A = Maximum of the values of Z from Curve Set I is
max(6, –24) = 6.

 

B = Maximum of the values of Z from Curve Set II is
max(11, –1, –18) = 11.

 

C = Minimum value of the values of Z from Curve Set I is min (6, –24) = –24.

 

D = Minimum of the values of Z from Curve Set II =
min(11, –1, –18) = –18.

 

E = the Maximum of the values of A, B, C, and D above + the Minimum of the values A, B, C, and D above = 11 + (–24) = –13

 

The result is
 
Maximum Minimum  
6
11
-24
-18
-13
  1. We know that,
A = Maximum of Z from Curve Set I.
 
B = Maximum of Z from Curve Set II.
 
C = Minimum of Z from Curve Set I and therefore smaller than A, unless both are equal.
 
D = Minimum of Z from Curve Set II and therefore smaller than B, unless both are equal.
 
Now,
 
E = (Maximum of the values of A, B, C, and D) + (the Minimum of the values of A, B, C, and D)
= (A or B whichever is larger) + (C or D whichever is smaller)
 
And
 
F = (Maximum of the values of A, B, C, and D) – (the Minimum of the values of A, B, C, and D)
= (A or B whichever is larger) – (C or D whichever is smaller)
 
Say, E = Maximum + Minimum and F = MaximumMinimum.
 
Column I:
 
Column I says that C is negative which indicates Minimum is negative. Now, E = Maximum + Negative and F = MaximumNegative = Maximum + Positive > Maximum. Therefore, F is greater than the rest. Select Row 5 in Column I of the answer table.
 
Column II:
 
We are asked to treat columns separately. Hence, the information, or the corresponding derivations, in Column I are not known here.
 
Now, if E is greater than F, then
Maximum + Minimum > MaximumMinimum
Minimum > – Minimum (by cancelling Maximum from both sides of the inequality)
2 x Minimum > 0
Or Minimum is positive.
 
Hence, either the minimum of C and D is positive or both C and D are positive.
 
Now, since A and B are greater than C and D and since C is positive, A and B must be positive.
 
Also, E is greater than the rest of A, B, C, D, and F because Minimum is positive. Note: C could still be positive here unlike in Column I.
 
Choose correspondingly.
 
The result should look like this:
 
C is negative E is greater than F  
A is positive.
B is positive.
D is positive.
E is greater than ABCD, and F.
F is greater than ABCD, and E.




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