A company manufactures two products, \(A\) and \(B\), on two machines, \(\bar{I}\) and II. It has been determined that the company will realize a profit of $$\$ 3$$ on each unit of product \(A\) and a profit of $$\$ 4$$ on each unit of product \(\mathrm{B}\). To manufacture a unit of product A requires $6 \mathrm{~min}\( on machine \)\mathrm{I}\( and \)5 \mathrm{~min}$ on machine II. To manufacture a unit of product B requires 9 min on machine \(\mathrm{I}\) and $4 \mathrm{~min}\( on machine \)\mathrm{II}\(. There are \)5 \mathrm{hr}$ of machine time available on machine \(\mathrm{I}\) and \(3 \mathrm{hr}\) of machine time available on machine II in each work shift. How many units of each product should be produced in each shift to maximize the company's profit?

Solve each linear programming problem by the method of corners. $$ \begin{array}{cc} \text { Maximize } & P=3 x-4 y \\ \text { subject to } & x+3 y \leq 15 \\ & 4 x+y \leq 16 \\ & x \geq 0, y \geq 0 \end{array} $$

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. Choosing the pivot column by requiring that it be the column associated with the most negative entry to the left of the vertical line in the last row of the simplex tableau ensures that the iteration will result in the greatest increase or, at worse, no decrease in the objective function.

Solve each linear programming problem by the simplex method. $$ \begin{array}{ll} \text { Maximize } & P=5 x+4 y \\ \text { subject to } & 3 x+5 y \leq 78 \\ & 4 x+y \leq 36 \\ & x \geq 0, y \geq 0 \end{array} $$

Determine whether the given simplex tableau is in final form. If so, find the solution to the associated regular linear programming problem. If not, find the pivot element to be used in the next iteration of the simplex method. $$ \begin{array}{rrrrr|r} x & y & u & v & P & \text { Constant } \\ \hline 0 & \frac{1}{2} & 1 & -\frac{1}{2} & 0 & 2 \\ 1 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & 4 \\ \hline 0 & -\frac{1}{2} & 0 & \frac{3}{2} & 1 & 12 \end{array} $$

Use the technique developed in this section to solve the minimization problem. $$ \begin{aligned} \text { Minimize } & C=-2 x+y \\ \text { subject to } & x+2 y \leq 6 \\ & 3 x+2 y \leq 12 \\ & x \geq 0, y \geq 0 \end{aligned} $$