For the LPP,
\( Z=20x+10y \)
Subject to
\( x+2y\le 40 \)
\( 3x+y\ge 30 \)
\( 4x+3y\ge 60 \)
\( x,y\ge 0 \)
Match List-I with List-II
| List-I | List-II | ||
| (A) | \( \text{Max. }Z \) | (I) | 180 |
| (B) | \( \text{Min. }Z \) | (II) | 320 |
| (C) | \( \text{Max. }Z-2\,\text{Min. }Z \) | (III) | 240 |
| (D) | If \( \text{Min. }Z \) is attained at \( (\alpha,\beta) \), then \( \alpha^2+\beta^2 \) is | (IV) | 800 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (IV), (B) - (III), (C) - (II), (D) - (I)
The feasible region lies below \( x+2y=40 \) and above both \( 3x+y=30 \) and \( 4x+3y=60 \), in the first quadrant. Its corner points are:
\( \displaystyle x+2y=40\ \text{and}\ 3x+y=30:\ (4,\ 18) \)
\( \displaystyle 3x+y=30\ \text{and}\ 4x+3y=60:\ (6,\ 12) \)
\( \displaystyle 4x+3y=60\ \text{and}\ y=0:\ (15,\ 0) \)
\( \displaystyle x+2y=40\ \text{and}\ y=0:\ (40,\ 0) \)
On the y-axis the constraints need \( y\ge 30 \) and \( y\le 20 \) together, so no corner lies there; for example (0, 20), where \( x+2y=40 \) meets \( 4x+3y=60 \), fails \( 3x+y\ge 30 \). The region is bounded, with these four corners.
\( \displaystyle Z(4,18)=80+180=260 \)
\( \displaystyle Z(6,12)=120+120=240 \)
\( \displaystyle Z(15,0)=300 \)
\( \displaystyle Z(40,0)=800 \)
(A) Max. Z
This matches (IV) 800, at (40, 0).
(B) Min. Z
This matches (III) 240, at (6, 12).
(C) Max. Z − 2 Min. Z
This matches (II) 320.
\( \displaystyle 800-2\times 240=320 \)
(D) \( \alpha^2+\beta^2 \) at the minimum point \( (\alpha,\beta)=(6,12) \)
This matches (I) 180.
\( \displaystyle 6^2+12^2=36+144=180 \)
So the matching is (A)-(IV), (B)-(III), (C)-(II), (D)-(I).