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Worked Examples · Example 6

Q.Solve the following Linear Programming Problem graphically. Maximize Z=5x+3yZ = 5x + 3y Subject to constraints:
3x+5y≤153x + 5y \le 15
5x+2y≤105x + 2y \le 10
x≥0, and y≥0x \ge 0, \text{ and } y \ge 0

Sikkim CbseNCERTSubjective· 5mImportance★★★★★
27% · 6/22 Questions
✓ Free question

The maximum of Z=5x+3yZ = 5x + 3y over the feasible region is 23519≈12.37\dfrac{235}{19} \approx 12.37, attained at (2019,4519)\left(\tfrac{20}{19}, \tfrac{45}{19}\right).

Corner points of the feasible region

With 3x+5y≤153x+5y\le 15, 5x+2y≤105x+2y\le 10, x,y≥0x,y\ge 0 (testing the origin in both, the region lies toward the origin), the feasible region has vertices:

  • (0,0)(0,0)
  • (2,0)(2,0) — where 5x+2y=105x+2y=10 meets the xx-axis
  • (0,3)(0,3) — where 3x+5y=153x+5y=15 meets the yy-axis
  • Intersection of 3x+5y=153x+5y=15 and 5x+2y=105x+2y=10:

15x+25y=75,15x+6y=30  ⇒  19y=45,  y=451915x+25y=75,\quad 15x+6y=30 \;\Rightarrow\; 19y=45,\; y=\tfrac{45}{19}

5x=10−9019=10019  ⇒  x=20195x = 10 - \tfrac{90}{19} = \tfrac{100}{19} \;\Rightarrow\; x=\tfrac{20}{19}

giving (2019,4519)\left(\tfrac{20}{19}, \tfrac{45}{19}\right).

Evaluate Z=5x+3yZ = 5x + 3y

CornerZZ
(0,0)(0,0)00
(0,3)(0,3)99
(2,0)(2,0)1010
(2019,4519)\left(\tfrac{20}{19}, \tfrac{45}{19}\right)100+13519=23519≈12.37\dfrac{100+135}{19} = \dfrac{235}{19} \approx 12.37

The largest value is 23519\dfrac{235}{19}.

Figure 8.1 — feasible region OABC for Maximize Z = 5x + 3y subject to 3x+5y<=15 and 5x+2y<=10
Figure 8.1 — feasible region OABC for Maximize Z = 5x + 3y subject to 3x+5y<=15 and 5x+2y<=10

The bounded feasible region OABC; the maximum of Z occurs at corner B(20/19, 45/19).

✓Final answer

The maximum value is Z=23519≈12.37Z = \dfrac{235}{19} \approx 12.37, occurring at (2019,4519)\left(\dfrac{20}{19}, \dfrac{45}{19}\right).

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