Please answer asap. WIll upvote as well. Thanks

A farmer must decide what crops to grow on a 300-hectare tract of land. He can grow oats, wheat or barley, which yield 50, 100 and 80 kg/hectare respectively, and sell for $ 1.00, $, 0.80 and $0.60 per kg respectively. Production costs (fertilizer, labour, etc.) are $40, $50 and $40 per hectare for growing oats, wheat and barley, respectively. Government regulations restrict the farmer to a maximum of 150 hectares of wheat and his crop rotation schedule requires that he plant at least 60 hectares in oats and 70 hectares in barley. Because of his storage arrangements, the farmer wants the number of hectares of oats to be equal to or less than half the number of hectares of barley. The farmer would like to maximize profit.

Let X1, X2 and X3 be the number of hectares planted in oats, wheat and barley respectively. Some useful information can be computed as follows:

The questions below are independent of each other. When answering any one of them, ignore the others. Answer all the questions below ACCORDING TO the above reports.

1. What are the optimal values of the decision variables? What is the optimal value of the objective function? (3 points)

2. The Government regulations changed the restrictions to a maximum of 130 hectares of wheat, does this impact the optimal profits? Justify your answer. (3 points)

3. The profit per hectare of wheat has decreased to $28. How does this affect the optimal values of the decision variables and the profits? Justify your answer. (5 points)

4. The number of hectares to be planted has increased to 320, and the minimum required area for oats has increased to 70. Management would like your advice on how these changes will affect profits. Justify your answer. (6 points)

5. The selling price per kg for oats has increased to $ 1.20. How does this affect the optimal solution? Justify your answer. (5 points)

revenue cost per kg produced per hectare profit per hectare Product Hectares per hectare hectare Oats X1 50*X1 50*X1 40*x1 10 Wheat X2 100*X2 80*X2 50*X2 30 Barley X3 80*X3 48*X3 40*X3 8 The tables below are copied from the SOLVER output. 60 120 120 Hectares Hectares Hectares of Oats of Wheat of Barley Profit Constraints 10 30 8 5160 1 1 1 300 300 1 0 0 60 60 Total to plant Minimum area oats Maximum area wheat Minimum area barley Oats-barley ratio 0 1 0 120 =$H$27 Binding 120$F$28=$H$29 Not Binding O$F$30=$H$27 Binding 120$F$28=$H$29 Not Binding O$F$30