A local bakery sells cakes and pies. The cakes take 2 hours of preparation time and 1 hour to decorate. The pies take 3 hours of preparation time and 0.5 hours to decorate. The profit is $25 per cake and $12 per pie. The bakery can spend at most 108 hours on preparation and 20 hours on decorating their items per day.
a.Write the system of linear inequalities to represent this situation.
b.Sketch the feasible region.
c.Write the optimization equation.
d.Determine the maximum profit. Show all work.
a) Let's assume C represents the number of cakes sold and P represents the number of pies sold.
The inequality for preparation time is:
2C + 3P ≤ 108 (since cakes take 2 hours and pies take 3 hours to prepare)
The inequality for decoration time is:
1C + 0.5P ≤ 20 (since cakes take 1 hour and pies take 0.5 hours to decorate)
The non-negativity constraints are:
C ≥ 0 (the number of cakes cannot be negative)
P ≥ 0 (the number of pies cannot be negative)
b) To sketch the feasible region, we need to plot the graph of each inequality and shade the area where the inequalities are satisfied.
For the preparation time:
2C + 3P ≤ 108
We can rewrite this as:
P ≤ (108 - 2C) / 3
For the decoration time:
1C + 0.5P ≤ 20
We can rewrite this as:
P ≤ 40 - 2C
Now, let's plot the graphs:
First, draw the line P = (108 - 2C) / 3.
Then draw the line P = 40 - 2C.
The feasible region is the region where both inequalities are satisfied, which is below both lines.
c) The optimization equation is the equation that represents the objective we want to maximize. In this case, the objective is to maximize profit. The profit equation can be written as:
Profit = 25C + 12P
d) To determine the maximum profit, we need to find the corner point within the feasible region that maximizes the profit equation.
Let's find the intersection of the two lines:
(108 - 2C) / 3 = 40 - 2C
108 - 2C = 3(40 - 2C)
108 - 2C = 120 - 6C
4C = 12
C = 3
Substituting C = 3 into P = (108 - 2C) / 3:
P = (108 - 2(3)) / 3
P = (108 - 6) / 3
P = 102 / 3
P = 34
The intersection point is (C, P) = (3, 34).
Now, substitute these values into the profit equation:
Profit = 25C + 12P
Profit = 25(3) + 12(34)
Profit = 75 + 408
Profit = 483
So, the maximum profit is $483.
a. To represent this situation as a system of linear inequalities, let's define the following variables:
- Let x represent the number of cakes produced.
- Let y represent the number of pies produced.
The time constraints can be represented by the following inequalities:
- Preparation time constraint: 2x + 3y ≤ 108 hours
- Decorating time constraint: x + 0.5y ≤ 20 hours
The profit can be calculated using the following equation:
- Profit equation: P = 25x + 12y
b. To sketch the feasible region, we need to graph the inequalities mentioned above. Let's start with the preparation time constraint, 2x + 3y ≤ 108:
To graph this inequality, we can rewrite it in the form y ≤ (-2/3)x + 36. Let's plot this line:
The y-intercept is 36, and the slope is -2/3. Therefore, we start at (0, 36) and go down 2 units and right 3 units to plot the next point. We connect the points to draw a line. Since the inequality is y ≤ (-2/3)x + 36, the feasible region will be below this line.
Next, let's graph the decorating time constraint, x + 0.5y ≤ 20:
Simplifying this inequality, we get y ≤ -2x + 40. Let's plot this line:
The y-intercept is 40, and the slope is -2. We start at (0, 40) and go down 2 units and right 1 unit to plot the next point. We connect the points to draw a line. Since the inequality is y ≤ -2x + 40, the feasible region will be below this line as well.
The feasible region is the shaded area below both lines.
c. The optimization equation is the profit equation: P = 25x + 12y.
d. To determine the maximum profit, we need to find the point within the feasible region that maximizes the profit equation P = 25x + 12y.
This can be achieved by evaluating the profit equation at the vertices of the feasible region and finding the maximum value.
Let's find the coordinates of the vertices by solving the set of equations formed by the lines:
- Intersection point 1: Solve the system of equations 2x + 3y = 108 and x + 0.5y = 20 to find the first vertex point.
- Intersection point 2: Solve the system of equations 2x + 3y = 108 and y = 0 to find the second vertex point.
- Intersection point 3: Solve the system of equations x + 0.5y = 20 and y = 0 to find the third vertex point.
Evaluate the profit equation P = 25x + 12y at each of the vertex points and choose the maximum profit value.
After performing these steps, the maximum profit can be determined.