Question #269592

A retail store stocks two types of customized travel bags, A and B. The store can sell a maximum of 400 type of A bags and a maximum if 300 type B bags per week. However, the store can only store up to 600 bags of both types because of limited storage capacity. The store earnd the profit of 30 per type A bag and a profit of 50 per type B bag. How many of each type of bags should the store keep per week to maximize the profit?

1
Expert's answer
2021-11-22T19:56:28-0500

Let x1 represent bag of type a and x2 represent bag of type bThe linear program in the given problem is Maximize:30x1+50x2Subject to: 400x1+300x2600 The linear program in its standard form is Maximize:30x1+50x2+0x3Subject to: 400x1+300x2+x3=600Next we form our first Tableau from our linear programx1x2x330500x34003001600305000Next, we locate the most negative number in the bottom row(-7), and label the columnwhere it is found the work column , we then form positive ratios by dividing the elementsin the work column by corresponding elements in the last column. Next we label the the smallest positive ratio, the pivot element. Using elementary row operations we reduce the pivot element to 1 and other elements in the work column to 0 to obtainour tableau 2x1x2x330500x250431160011030030000Therefore 50 of x2 should be produced\displaystyle \text{Let $x_1$ represent bag of type a and $x_2$ represent bag of type b}\\ \text{The linear program in the given problem is }\\ Maximize: 30x_1 +50x_2\\ \text{Subject to: } 400x_1 +300x_2\leq 600\\\ \text{The linear program in its standard form is }\\ Maximize: 30x_1 +50x_2+0x_3\\ \text{Subject to: } 400x_1 +300x_2 +x_3 = 600\\ \text{Next we form our first Tableau from our linear program}\\ \begin{matrix} & & x_1 & x_2 & x_3 \\ & & 30 & 50 & 0 \\\hline x_3 & 400 & 300& 1 & 600 \\\hline & & -30 & -50 & 0 & 0 \\ \end{matrix}\\ \text{Next, we locate the most negative number in the bottom row(-7), and label the column}\\ \text{where it is found the work column , we then form positive ratios by dividing the elements}\\ \text{in the work column by corresponding elements in the last column. Next we label the }\\ \text{the smallest positive ratio, the pivot element. Using elementary row operations we }\\ \text{reduce the pivot element to 1 and other elements in the work column to 0 to obtain}\\ \text{our tableau 2}\\ \begin{matrix} & & x_1 & x_2 & x_3 \\ & & 30 & 50 & 0 \\\hline x_2 & 50 & \frac{4}{3} & 1 & 1 & 600 \\\hline && \frac{110}{3} & 0 &0 &30000\\ \end{matrix}\\ \text{Therefore 50 of $x_2$ should be produced}


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