Chapter 4, Problem 31CCSR

Solution

a.

To find: The system of inequalities of the situation.

  $\begin{array}{c}x+y\le 80\\ 0.10x+0.05y\le 6\\ x\ge 0,y\ge 0\end{array}$

Given: A man has at most 80 brownies and cookies. The cost of each brownie $0.10 and cost of each cookie $0.05.

Explanation:

Let be $x$ number of brownies and $y$ numbers of cookies.

There are at most 80 brownies and cookies.

  $x+y\le 80$

Total spend amount is $6.

Therefore, the inequality is $0.10x+0.05y\le 6$

  $x\ge 0,y\ge 0$

b.

To graph: The system of inequalities.

Given: A man has at most 80 brownies and cookies. The cost of each brownie $0.10 and cost of each cookie $0.05.

Explanation:

The system of inequalities,

  $\begin{array}{c}x+y\le 80\\ 0.10x+0.05y\le 6\\ x\ge 0,y\ge 0\end{array}$

The graph of the inequalities,

  

Now choose any one point of the feasible region. That point satisfy the both situation of the problem.

c.

To find: Each desert which maximize the profit.

40 and 40

Given: A man has at most 80 brownies and cookies. The cost of each brownie $0.10 and cost of each cookie $0.05. A profit of $0.25 on each brownie and $0.20 on each cookie.

Explanation:

The optimization function, $Z=0.25x+0.20y$

From the graph the corner points and find the profit.

  $\begin{array}{cccc}\text{Points}& Z=0.25x+0.20y& \text{Calculated}& \text{Conclusion}\\ \left(0,0\right)& Z=0.25\left(0\right)+0.20\left(0\right)& 0& \\ \left(0,80\right)& Z=0.25\left(0\right)+0.20\left(80\right)& 16& \\ \left(40,40\right)& Z=0.25\left(40\right)+0.20\left(40\right)& 18& \max \\ \left(60,0\right)& Z=0.25\left(60\right)+0.20\left(0\right)& 15& \end{array}$

The profit is maximum when 40 units of brownies and 40 units of cookies sale.