$1.500 om 1004 Apply the g Cka e kon Code Manahang pro red packs in which The cost per share and expected rett over the next two years is given in the accompanying table A But model Solver found that the optimal solution was to buy 115 38 shares of stock A0 shares of stock and 9 shares of stock C, which results in a re Cho Manufacturing model to find the spinal son Compare your aner with the Solver solution Top by ning the same hd as the Ondo Manufacturing problem ist find the catio of the the free sacks The res Found decimal places an for stock C coficient to the coefficient for each of Crebo Manufacturing produces four types of structural support fittings plugs, rails, rivets, and clips-which are machined on two CNC machining centers. The machining centers have a capacity of 280,000 minutes per year. The gross margin per unit and machining requirements are provided in the table below. Stock Price/share Return/share A B C $13 $15 $35 $9 $5 $11 How many of each product should be made to maximize gross profit margin? To formulate this as a linear optimization model, define X,, X2, X3, and X4 to be the number of plugs, rails, rivets, and clips, respectively, to produce. The problem is to maximize gross margin = 0.3X, +1.3X2 +0.75X3 + 1.2X4 subject to the constraint that limits the machining capacity and nonnegativity of the variables: 1X, +2.5 X+1.5 X + 2 X 280,000 X1, X2, X3, X420 To solve this problem, your first thought might be to choose the variable with the highest marginal profit. Because X2 has the highest marginal profit, you might try producing as many rails as possible. Since each rail requires 2.5 minutes. constraint that limits the machining capacity and nonnegativity of the variables: 1X, +2.5 X+1.5X3+2X4280,000 X1, X2, X3, X20 To solve this problem, your first thought might be to choose the variable with the highest marginal profit. Because X2 has the highest marginal profit, you might try producing as many rails as possible. Since each rail requires 2.5 minutes, the maximum number that can be produced is 280,000 > 2.5= 112,000, for a total profit of $1.3(112,000) = $145,600. However, notice that each rail uses a lot more machining time than the other products. The best solution isn't necessarily the one with the highest marginal profit, but the one that provides the highest total profit. Therefore, more profit might be realized by producing a proportionately larger quantity of a different product having a smaller marginal profit. This is the key insight. What the simplex method essentially does is evaluate the impact of constraints in terms of their contribution to the objective function for each variable. For the simple case of only one constraint, the optimal (maximum) solution is found by simply choosing the variable with the highest ratio of the objective coefficient to the constraint coefficient. The highest ratio occurs for clips. If we produce the maximum number of clips, 280,000/2=140,000, the total profit is $1.20 (140,000) = $168,000