A local shop sells two different types of tea. They know that the maximum number of cups of orange pekoe tea and chamomile tea they sell in a day is 250. For a perfect cup of tea, the orange pekoe takes 15 minutes to brew and the chamomile takes 10 minutes to brew. Their brewing system is only available for 3000 minutes per day. Orange pekoe tea sells for $2.50 per cup and chamomile sells for $3.00 per cup.
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 x be the number of cups of orange pekoe tea sold, and y be the number of cups of chamomile tea sold.
The constraints are:
x ≥ 0 (non-negativity constraint)
y ≥ 0 (non-negativity constraint)
x + y ≤ 250 (maximum number of cups sold constraint)
15x + 10y ≤ 3000 (brewing system time constraint)
The objective function is:
Profit = $2.50x + $3.00y
b. The feasible region represents the area where all the constraints are satisfied. It can be graphed on a coordinate plane.
c. The optimization equation is to maximize the profit:
Maximize: Profit = $2.50x + $3.00y
d. To find the maximum profit, we need to evaluate the objective function at each corner point of the feasible region and determine the largest value.
The corner points are:
(0, 0)
(0, 250)
(200, 50)
(250, 0)
Evaluating the objective function:
Profit1 = $2.50(0) + $3.00(0) = $0
Profit2 = $2.50(0) + $3.00(250) = $750
Profit3 = $2.50(200) + $3.00(50) = $650
Profit4 = $2.50(250) + $3.00(0) = $625
The maximum profit is $750, which occurs when 0 cups of orange pekoe tea are sold and 250 cups of chamomile tea are sold.
a. Let's define the variables:
- Let x be the number of cups of orange pekoe tea sold.
- Let y be the number of cups of chamomile tea sold.
The constraints are:
1. The maximum number of cups sold in a day is 250:
x + y ≤ 250
2. The brewing time constraint is 15x + 10y ≤ 3000, because the brewing system can only be used for a total of 3000 minutes.
b. To sketch the feasible region, we need to graph the inequalities.
First, graph the equality x + y = 250. This is a straight line passing through points (0, 250) and (250, 0).
Next, graph the inequality 15x + 10y ≤ 3000. To do this, first draw the line 15x + 10y = 3000. To find the points on this line, you can choose any two x-values and solve for y. For example, when x = 0, y = 300, and when x = 200, y = 0. Connect the points (0, 300) and (200, 0) to define the line. Finally, shade the region below this line because we have the less than or equal to inequality.
c. The optimization equation is the objective function we want to maximize or minimize. In this case, we want to maximize profit.
Let P(x, y) be the profit function. We can express the profit as the total revenue minus the total cost:
P(x, y) = (2.50x + 3.00y) - 0
Since we have no other costs involved in this problem.
d. To determine the maximum profit, we need to find the corner points of the feasible region and substitute those values into the profit function.
The corner points of the feasible region are the intersections of the two lines:
(0, 250)
(200, 50)
(250, 0)
Substitute these points into the profit function:
P(0, 250) = (2.50 * 0 + 3.00 * 250) - 0 = $750
P(200, 50) = (2.50 * 200 + 3.00 * 50) - 0 = $650
P(250, 0) = (2.50 * 250 + 3.00 * 0) - 0 = $625
The maximum profit is $750.