Numerical Methods and Optimization

296 Numerical Methods and Optimization: An Introduction

then for any feasible solution we have:

z = 800x

1

+ 600x

2

= (250x

1

+ 450x

2

)1.1 + (250x

1

+50x

2

)2.1

≤ (9, 000 + p

1

)1.1+5, 000 · 2.1

=20, 400 + 1.1p

1

.

Thus, we got the following upper bound

z ≤ 20, 400 + 1.1p

1

on the objective, in which the optimal value of the first dual variable y

∗

1

=1.1

is the coefficient for the variable representing the extra Merlot grapes. The

extra profit added to the currently optimal profit of $20,400 will never exceed

1.1p

1

, thus $1.1/pound is the maximum extra amount the winery should be

willing to pay for additional Merlot grapes.

Similarly, if the winery looks to purchase p

2

more