Chapter 7 - Algebra: Graphs, Functions, and Linear Systems - Chapter Summary, Review, and Test - Review Exercises - Page 482: 64

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(a) Take, \[x\] as the number of units of writing paper produced daily further take \[y\]as the number of units of newsprint produced daily. So, the profit of unit writing produced daily is \[500x\] and the profit of newsprint produced daily is\[350y\]. Take, \[z\]to represent total daily profit. So, the objective function is: \[z=500x+350y\] Hence, the objective function that models total daily profit is written below: \[z=500x+350y\]. (b) Take, \[x\] as the number of units of writing paper produced daily. Further, take \[y\]as the number of units of newsprint produced daily. As the factory allows at most \[200\]units of paper in a day, then the constraints are: \[x+y\le 200\] The customer requires at least \[10\]units of writing paper, then the constraints are: \[x\ge 10\] Further customers require at least \[80\]units of newsprint daily, then the constraints are: \[y\ge 80\] Hence, the constraints are as follows: \[\left\{ \begin{align} & x+y\le 200, \\ & x\ge 10, \\ & y\ge 80 \\ \end{align} \right\}\] (d) Consider the objective function from part (a): \[z=500x+350y\] Substitute, all the three vertices in the objective function to get the total profit which is written below: \[\left( 10,190 \right),\left( 10,80 \right),\left( 120,80 \right)\] So, start with the first one that is \[\left( 10,190 \right)\]. \[\begin{align} & z=500x+350y \\ & =500\left( 10 \right)+350\left( 190 \right) \\ & =5000+66500 \\ & =71500 \end{align}\] Then, the second vertices are \[\left( 10,80 \right)\]. \[\begin{align} & z=500x+350y \\ & =500\left( 10 \right)+350\left( 80 \right) \\ & =5000+28000 \\ & =33000 \end{align}\] Next, the last vertices are \[\left( 120,80 \right)\]. \[\begin{align} & z=500x+350y \\ & =500\left( 120 \right)+350\left( 80 \right) \\ & =60000+28000 \\ & =88000 \end{align}\] Hence, the objective function at \[\left( 10,190 \right)\] is\[71500\], \[\left( 10,80 \right)\] is \[33000\] and \[\left( 120,80 \right)\] is\[88000\]. (e) Substitute, all the three vertices in the objective function to get the total profit, which is written below: \[\left( 10,190 \right),\left( 10,80 \right),\left( 120,80 \right)\] So, start with the first one, which is \[\left( 10,190 \right)\]. \[\begin{align} & z=500x+350y \\ & =500\left( 10 \right)+350\left( 190 \right) \\ & =5000+66500 \\ & =71500 \end{align}\] Then, the second one vertices are \[\left( 10,80 \right)\] \[\begin{align} & z=500x+350y \\ & =500\left( 10 \right)+350\left( 80 \right) \\ & =5000+28000 \\ & =33000 \end{align}\] Next, the last one vertices are \[\left( 120,80 \right)\] \[\begin{align} & z=500x+350y \\ & =500\left( 120 \right)+350\left( 80 \right) \\ & =60000+28000 \\ & =88000 \end{align}\] To maximize the profit, then consider that point which gives the maximum value, so the maximum daily profit is \[\$88000\]and the vertices are x ..