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Maximise the function Z = 11x + 7y, subject to the constraints: x ≤ 3, y ≤ 2, x ≥ 0, y ≥ 0.

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प्रश्न

Maximise the function Z = 11x + 7y, subject to the constraints: x ≤ 3, y ≤ 2, x ≥ 0, y ≥ 0.

योग
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उत्तर

The shaded region is the feasible region determined by the constraints x ≤ 3, y ≤ 2, x ≥ 0, y ≥.

The feasible region is bounded with four corners O(0, 0), A(3, 0), B(3, 2) and C(0, 2).

So, the maximum value can occur at any corner

Let us evaluate the value of Z.

Corner points Value of Z  
O(0, 0) 11(0) + 7(0) = 0  
A(3, 0) 11(3) + 7(0) = 33  
B(3, 2) 11(3) + 7(2) = 47 ← Maximum
C(0, 2) 11(0) + 7(2) = 14  

Hence, the maximum value of the function Z is 47 at (3, 2).

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अध्याय 12: Linear Programming - Exercise [पृष्ठ २५०]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 12 Linear Programming
Exercise | Q 3 | पृष्ठ २५०

वीडियो ट्यूटोरियलVIEW ALL [2]

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