Question Details

Maximize Z = 3x + 5y, subject to x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0

Options

A

20 at (1, 0)

B

30 at (0, 6)

C

37 at (4, 5)

D

33 at (6, 3)

Correct Answer :

37 at (4, 5)

Solution :

The correct option is 37 at (4, 5).

To find the maximum value of the objective function Z=3x+5y, we first determine the feasible region defined by the following system of linear inequalities:
1) x+4y24
2) 3x+y21
3) x+y9
4) x0,y0

The non-negativity constraints (x0,y0) restrict the feasible region to the first quadrant. We find the boundary lines corresponding to the inequalities:
Line 1: x+4y=24. The intercepts are (24, 0) and (0, 6).
Line 2: 3x+y=21. The intercepts are (7, 0) and (0, 21).
Line 3: x+y=9. The intercepts are (9, 0) and (0, 9).

Next, we identify the corner points of the bounded feasible region by finding the intersection points of these lines:
- The origin (0,0) is a corner point.
- The y-intercept of the feasible region on the y-axis is determined by Line 1, which gives (0,6) (since 69 and 621).
- The x-intercept of the feasible region on the x-axis is determined by Line 2, which gives (7,0) (since 79 and 724).

Now we find the intersection points of the lines in the first quadrant:
- Intersection of Line 1 (x+4y=24) and Line 3 (x+y=9):
Subtracting the second equation from the first gives:
3y=15y=5.
Substituting y=5 into x+y=9 gives x=4.
This yields the intersection point (4, 5). We check if (4, 5) satisfies Line 2's inequality: 3(4)+5=1721, which is true. Thus, (4, 5) is a valid corner point.

- Intersection of Line 2 (3x+y=21) and Line 3 (x+y=9):
Subtracting the second equation from the first gives:
2x=12x=6.
Substituting x=6 into x+y=9 gives y=3.
This yields the intersection point (6, 3). We check if (6, 3) satisfies Line 1's inequality: 6+4(3)=1824, which is true. Thus, (6, 3) is also a valid corner point.

Now, we evaluate the objective function Z=3x+5y at each of the corner points of the feasible region:
1. At (0,0): Z=3(0)+5(0)=0
2. At (7,0): Z=3(7)+5(0)=21
3. At (6,3): Z=3(6)+5(3)=18+15=33
4. At (4,5): Z=3(4)+5(5)=12+25=37
5. At (0,6): Z=3(0)+5(6)=30

Comparing the values, the maximum value of Z is 37, which occurs at the corner point (4,5).

Unlock Our Free Library

Access expert-curated educational resources and study materials—completely free.

Discover more resources

You may also like

Mock Tests

View All
  • JEE
  • intermediate
  • 3 hours
  • chemistry, mathematics, physics

  • JEE
  • intermediate
  • 3 hours
  • chemical engineering, mathematics, physics