Problem-4 (Multi-Source-Integration between Curves)
Curve Set I |
Curve Set II |
Curve Set III |
Curve 1: Curve 2: |
Curve 3: Curve 4: |
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 â€“ the Minimum of the values of A, B, C, and D above. |
- If x = â€“3 and y = 4, then identify the expressions whose values are maximum or minimum?
Maximum | Minimum | |
A | ||
B | ||
C | ||
D | ||
E |
- 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)
- Substituting x = â€“3 and y = 4 yields
Curve Set I |
Curve Set II |
Curve Set III |
Curve 1: Z = 2x + 3y
Curve 2: Z = 4x â€“ 3y |
Curve 3: Z = 2y â€“ x
Curve 4: Z = 2y + 3x
Curve 5: Z = 2x â€“ 3y |
Curves A, B, C, and, D
A = Maximum of the values of Z from Curve Set I is
B = Maximum of the values of Z from Curve Set II is
C = Minimum value of the values of Z from Curve Set I is min (6, â€“24) = â€“24.
D = Minimum of the values of Z from Curve Set II =
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 + (â€“24) = â€“13 |
Maximum | Minimum | |
6 | ||
11 | ||
-24 | ||
-18 | ||
-13 |
- We know that,
Maximum + Minimum > Maximum â€“Minimum
Minimum > â€“ Minimum (by cancelling Maximum from both sides of the inequality)
2 x Minimum > 0
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. |