Solve the following Linear Programming Problem graphically :
Maximize \( Z = 8x + 8y \)
subject to the constraints
\[ \begin{aligned} x - 20 &\leq 0 \\ 2x + 3y &\leq 120 \\ 2x + y &\leq 60 \\ x \geq 0,\ y &\geq 0 \end{aligned} \]
Solve the following Linear Programming Problem graphically :
Maximize \( Z = 8x + 8y \)
subject to the constraints
\[ \begin{aligned} x - 20 &\leq 0 \\ 2x + 3y &\leq 120 \\ 2x + y &\leq 60 \\ x \geq 0,\ y &\geq 0 \end{aligned} \]
Show Hint
Solution and Explanation
Maximise \( Z = 8x + 8y \) subject to \( x \leq 20 \), \( 2x + 3y \leq 120 \), \( 2x + y \leq 60 \), \( x, y \geq 0 \).
Draw the boundary lines: \( x = 20 \); \( 2x + 3y = 120 \) through \( (0, 40) \) and \( (60, 0) \); \( 2x + y = 60 \) through \( (0, 60) \) and \( (30, 0) \).
Step 2: Decide the feasible region:
The origin satisfies all three inequalities, so the region lies on the origin side of each line. The condition \( x, y \geq 0 \) keeps us in the first quadrant.
The feasible region is a bounded polygon with five corners.
Step 3: Find the corner points:
The corner \( O \) is the origin \( (0, 0) \). On the x-axis, the line \( x = 20 \) cuts \( y = 0 \) at \( A(20, 0) \).
Put \( x = 20 \) in \( 2x + y = 60 \). Then \( y = 20 \), so \( B(20, 20) \). This point also satisfies \( 2x + 3y = 40 + 60 = 100 \leq 120 \).
Solve \( 2x + 3y = 120 \) and \( 2x + y = 60 \). Subtract to get \( 2y = 60 \), so \( y = 30 \) and \( x = 15 \). This gives \( C(15, 30) \).
The line \( 2x + 3y = 120 \) meets the y-axis at \( D(0, 40) \).
The line \( x = 20 \) meets \( 2x + 3y = 120 \) at \( \left(20, \tfrac{80}{3}\right) \). This point breaks \( 2x + y \leq 60 \), because \( 40 + \tfrac{80}{3} \gt 60 \). So it is not a corner.
So the corners are \( O(0, 0),\ A(20, 0),\ B(20, 20),\ C(15, 30),\ D(0, 40) \).
Step 4: Evaluate Z at each corner:
A bounded region gives its maximum at a corner point.
\( Z(O) = 0 \)
\( Z(A) = 8(20) + 8(0) = 160 \)
\( Z(B) = 8(20) + 8(20) = 320 \)
\( Z(C) = 8(15) + 8(30) = 120 + 240 = 360 \)
\( Z(D) = 8(0) + 8(40) = 320 \)
Final Answer:
The largest value of \( Z \) is 360, reached at \( x = 15,\ y = 30 \). \[ \boxed{Z_{\max} = 360 \text{ at } (15, 30)} \]
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