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📅 May 8, 2026 👤 scraface

May 08, 2026

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2x + 3y \leq 120
and non-negativity:
x \geq 0, \quad y \geq 0
Solving using linear programming, find the intercepts of the constrain...
If \( x = 0 \), \( 3y = 120 \rightarrow y = 40 \)
If \( y = 0 \), \( 2x = 120 \rightarrow x = 60 \)
The feasible region is bounded by the line and axes. Evaluate revenue...
At \( (0, 40) \): \( R = 50(0) + 80(40) = 3200 \)
At \( (60, 0) \): \( R = 50(60) + 80(0) = 3000 \)
Also, check the point where the line intersects the axes maximally:

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