9th Mathematics

Step 1: Graph the constraints in first quadrant:
• Line 1: 2y - x = 8. Intercepts are (-8, 0) and (0, 4). Shading is below (origin-side, since 0 ≤ 8).
• Line 2: x - y = 4. Intercepts are (4, 0) and (0, -4). Shading is above (origin-side, since 0 ≤ 4).

Step 2: Find corner points (vertices) of the feasible region:
• Origin: O(0, 0).
• x-intercept of Line 2: C(4, 0).
• y-intercept of Line 1: B(0, 4).
• Intersection point E of Line 1 and Line 2: Substitute into Line 2: So intersection point is E(16, 12).

Step 3: Evaluate objective function at corner points:

Corner Point (x, y)	f(x, y) = 2x + 5y	Value
O(0, 0)	2(0) + 5(0)	0
C(4, 0)	2(4) + 5(0)	8
B(0, 4)	2(0) + 5(4)	20
E(16, 12) [Optimal]	2(16) + 5(12) = 32 + 60	92

The maximum value is 92 at corner point E(16, 12).

Step 1: Graph constraints:
• Line 1: 2x + 5y = 30 (intercepts (15, 0), (0, 6)).
• Line 2: 5x + 4y = 20 (intercepts (4, 0), (0, 5)).

Step 2: Feasibility check:
Notice that Line 2 is completely inside the boundary of Line 1 in the first quadrant.
At point (0, 5), 2(0) + 5(5) = 25 ≤ 30 (True).
At point (4, 0), 2(4) + 5(0) = 8 ≤ 30 (True).
Thus, the intersection point of the boundary lines is outside the first quadrant, and the active feasible region is bounded by the axes and Line 2.

Step 3: Evaluate corner points:

Corner Point (x, y)	f(x, y) = x + 3y	Value
O(0, 0)	0 + 3(0)	0
C(4, 0)	4 + 3(0)	4
D(0, 5) [Optimal]	0 + 3(5)	15

The maximum value is 15 at corner point D(0, 5).

Step 1: Graph constraints:
• Line 1: 2x + y = 4 (intercepts (2, 0), (0, 4)).
• Line 2: 4x - y = 4 (intercepts (1, 0), (0, -4)).

Step 2: Find intersection E of both lines: Add the equations: Substitute x = 4/3: So intersection point is E(4/3, 4/3).

Step 3: Evaluate corner points:

Corner Point (x, y)	z = 2x + 3y	Value
O(0, 0)	2(0) + 3(0)	0
C(1, 0)	2(1) + 3(0)	2
E(4/3, 4/3)	2(4/3) + 3(4/3) = 8/3 + 4	20/3 ≈ 6.67
B(0, 4) [Optimal]	2(0) + 3(4)	12

The maximum value is 12 at corner point B(0, 4).

Step 1: Graph constraints:
• Line 1: x + y = 3 (intercepts (3, 0), (0, 3)). Shading is above (away from origin).
• Line 2: 7x + 5y = 35 (intercepts (5, 0), (0, 7)). Shading is below (towards origin).

Step 2: Identify corner points of feasible region:
The region lies between the two boundary lines in the first quadrant:
• Point A(3, 0) (on x-axis)
• Point C(5, 0) (on x-axis)
• Point D(0, 7) (on y-axis)
• Point B(0, 3) (on y-axis)

Step 3: Evaluate objective function at corner points:

Corner Point (x, y)	z = 2x + y	Value
B(0, 3) [Optimal Min]	2(0) + 3	3
A(3, 0)	2(3) + 0	6
D(0, 7)	2(0) + 7	7
C(5, 0)	2(5) + 0	10

The minimum value is 3 at corner point B(0, 3).

Step 1: Graph constraints:
• Line 1: 2x + y = 8 (intercepts (4, 0), (0, 8)).
• Line 2: x + 2y = 14 (intercepts (14, 0), (0, 7)).

Step 2: Find intersection E:
Multiply Line 1 by 2: Subtract Line 2 (x + 2y = 14) from it: Substitute x = 2/3 into Line 1: So intersection point is E(2/3, 20/3).

Step 3: Evaluate corner points:

Corner Point (x, y)	f(x, y) = 2x + 3y	Value
O(0, 0)	2(0) + 3(0)	0
A(4, 0)	2(4) + 3(0)	8
D(0, 7)	2(0) + 3(7)	21
E(2/3, 20/3) [Optimal]	2(2/3) + 3(20/3) = 4/3 + 20	64/3 ≈ 21.33

The maximum value is 64/3 ≈ 21.33 at corner point E(2/3, 20/3).

Step 1: Graph constraints:
• Line 1: 3x + 5y = 15 (intercepts (5, 0), (0, 3)). Shading is above (away from origin).
• Line 2: x + 6y = 9 (intercepts (9, 0), (0, 1.5)). Shading is above (away from origin).

Step 2: Find intersection point E: Multiply Line 2 by 3: Subtract equations: Substitute y = 12/13 into Line 2: So intersection point is E(45/13, 12/13).

Step 3: Evaluate corner points of unbounded region:

Corner Point (x, y)	z = 3x + y	Value
B(0, 3) [Optimal Min]	3(0) + 3	3
E(45/13, 12/13)	3(45/13) + 12/13 = 147/13	147/13 ≈ 11.31
C(9, 0)	3(9) + 0	27

The minimum value of z is 3 at corner point B(0, 3).