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Mathematics for Economy Question on Optimization

Consider a linear programming problem (๐‘ƒ) min ๐‘ง = 4๐‘ฅ1 + 6๐‘ฅ2 + 6๐‘ฅ3 subject to
๐‘ฅ1+3๐‘ฅ2โ‰ฅ3
๐‘ฅ1+2๐‘ฅ3 โ‰ฅ5
๐‘ฅ1, ๐‘ฅ2, ๐‘ฅ3 โ‰ฅ 0
If ๐‘ฅโˆ—=(๐‘ฅ1โˆ—,๐‘ฅ2โˆ—,๐‘ฅ3โˆ—)๐‘ฅ^โˆ— = (๐‘ฅ^โˆ—_1 , ๐‘ฅ^โˆ—_2 , ๐‘ฅ^โˆ—_3 ) is an optimal solution and ๐‘งโˆ— is an optimal value of (๐‘ƒ) and ๐‘คโˆ— =(๐‘ค1โˆ—,๐‘ค2โˆ—)(๐‘ค^โˆ—_1 , ๐‘ค^โˆ—_2 ) is an optimal solution of the dual of (๐‘ƒ) then

A

๐‘ฅ2โˆ—+๐‘ฅ3โˆ—=๐‘ค1โˆ—+๐‘ค2โˆ—๐‘ฅ^โˆ—_2 + ๐‘ฅ^โˆ—_3 = ๐‘ค^โˆ—_1 + ๐‘ค^โˆ—_2

B

๐‘งโˆ—=4(๐‘ฅ1โˆ—+๐‘ค2โˆ—)๐‘ง^โˆ— = 4(๐‘ฅ^โˆ—_1 + ๐‘ค^โˆ—_2 )

C

๐‘งโˆ—=6(๐‘ค1โˆ—+๐‘ฅ3โˆ—)๐‘ง^โˆ— = 6(๐‘ค^โˆ—_1 + ๐‘ฅ^โˆ—_3 )

D

๐‘ฅ1โˆ—+๐‘ฅ3โˆ—=๐‘ค1โˆ—+๐‘ค2โˆ—๐‘ฅ^โˆ—_1 + ๐‘ฅ^โˆ—_3 = ๐‘ค^โˆ—_1 + ๐‘ค^โˆ—_2

Answer

๐‘ฅ1โˆ—+๐‘ฅ3โˆ—=๐‘ค1โˆ—+๐‘ค2โˆ—๐‘ฅ^โˆ—_1 + ๐‘ฅ^โˆ—_3 = ๐‘ค^โˆ—_1 + ๐‘ค^โˆ—_2

Explanation

Solution

The correct option is (D): ๐‘ฅ1โˆ—+๐‘ฅ3โˆ—=๐‘ค1โˆ—+๐‘ค2โˆ—๐‘ฅ^โˆ—_1 + ๐‘ฅ^โˆ—_3 = ๐‘ค^โˆ—_1 + ๐‘ค^โˆ—_2