Question

Farmer Jones must determine how many acres of com and wheat to plant this year. An acre of wheat yields 25 bushels of wheat and requires 10 hours of labor per week. An acre of corn yields 10 bushels of corn and requires 4 hours of labor per week. All wheat can be sold at $4 a bushel, and all com can be sold at$3 a bushel. Seven acres of land and 40 hours per week of labor are available. Government regulations requite that at least 30 bushels of corn be produced during the current year. Let x1 = number of acres of corn planted, and x2 = number of acres of wheat planted. Using these decision variables, formulate an LP whose solution will tell Farmer Jones how to maximize the total revenue from wheat and corn.

Solution

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Decision variables\text{\underline{\color{#4257b2}Decision variables}}

The decision variables are already given.

Objective function\text{\underline{\color{#4257b2}Objective function}}

The profits need to be maximized. Since x1x_1 acres of wheat yields 25x125x_1 bushels of wheat, which sell for 44 dollars per bushel, so the revenue from x1x_1 acres of wheat is 425x1=100x14 \cdot 25x_1 = 100x_1 dollars.

Similarly, the revenue from x2x_2 acres of corn is 310x2=30x23 \cdot 10 x_2 = 30 x_2 dollars.

Therefore, the objective function is

max z=100x1+30x2\max \ z = 100x_1 + 30x_2

Constraints\text{\underline{\color{#4257b2}Constraints}}

Constraint 1:. Only 7 acres of land are available. Therefore, x1+x27x_1 + x_2 \leqslant 7. Constraint 2:. 40 hours of labor per week are available. Since x1x_1 acres of wheat need 10x110x_1 hours of labor per week, and x2x_2 acres of corn need 4x24x_2 hours of labor per week, we have 10x1+4x24010x_1 + 4x_2 \leqslant 40. Constraint 3:. The farmer needs to produce at least 30 bushels of corn. Since each acre of corn produces 1010 bushels of corn, we have 10x23010x_2 \geqslant 30, which can be simplified to x23x_2 \geqslant 3.

Sign restrictions\text{\underline{\color{#4257b2}Sign restrictions}}

Obviously, x10x_1 \geqslant 0 and x20x_2 \geqslant 0 (acres of corn and wheat must both be nonnegative).

Conclusion\text{\underline{\color{#4257b2}Conclusion}}

Finally, we can write the LP problem as

max z=100x1+30x2x1+0x20710x1+4x240x203x1+0x200x200\begin{align*} \max \ z &= 100x_1 + 30x_2 \tag{Objective function}\\ x_1 + \phantom{0}x_2 &\leqslant \phantom{0}7 \tag{Land restriction}\\ 10x_1 + 4x_2 &\leqslant 40 \tag{Labor restriction}\\ x_2 & \geqslant \phantom{0}3 \tag{Corn restriction}\\ x_1 \phantom{+0x_2} & \geqslant \phantom{0}0 \tag{Sign restriction} \\ x_2 & \geqslant \phantom{0}0 \tag{Sign restriction} \\ \end{align*}

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