Problem 76 - Precalculus

Question

A well-known donut store makes two popular types of donuts: crème-filled and jelly-filled. The manager knows from past statistics that the number of dozens of donuts sold is at least \(10,\) but no more than \(30 .\) To prepare the donuts for frying, the baker needs (on the average) 3 minutes for a dozen crème- filled and 2 minutes for jelly-filled. The baker has at most two hours available per day to prepare the donuts. How many dozens of each type should be prepared to maximize the daily profit if there is a \(\$ 1.20\) profit for each dozen crème-filled and \(\$ 1.80\) profit for each dozen jelly-filled donuts?

Step-by-Step Solution

Answer

Prepare 0 dozens of crème-filled and 30 dozens of jelly-filled donuts for maximum profit.

Step 1: Define the Variables

Let's define the variables for the problem. Let \( x \) be the number of dozens of crème-filled donuts, and \( y \) be the number of dozens of jelly-filled donuts. Our goal is to find the values of \( x \) and \( y \) that will maximize the daily profit.

Step 2: Formulate the Objective Function

The objective is to maximize the profit from selling donuts. The profit from a dozen crème-filled donuts is \( \\(1.20 \) and from a dozen jelly-filled donuts is \( \\)1.80 \). Thus, the objective function \( P \) for profit is: \[ P = 1.20x + 1.80y \] This is the function we aim to maximize.

Step 3: Establish Constraints

The problem includes several constraints based on the scenario description:1. The total number of dozens sold must be between 10 and 30: \[ 10 \leq x + y \leq 30 \]2. The total preparation time must not exceed 120 minutes (2 hours), with 3 minutes per dozen for crème-filled and 2 minutes per dozen for jelly-filled: \[ 3x + 2y \leq 120 \]3. The quantities \( x \) and \( y \) cannot be negative: \[ x \geq 0, \quad y \geq 0 \]

Step 4: Identify Feasible Region

Graph the inequalities in Step 3 to find the feasible region. This can be done by plotting each inequality, identifying where they overlap, and noting the boundary formed by each constraint. - The constraint \( x + y \geq 10 \) gives a region below and to the right of the line.- The constraint \( x + y \leq 30 \) gives a region above and to the left of the line.- The constraint \( 3x + 2y \leq 120 \) is a line that slants downward. The feasible region is where all these conditions are met simultaneously on a graph.

Step 5: Determine Corner Points

Identify the corner points of the feasible region using the intersection of the constraint lines. Solve the system of equations to find these points, including boundaries:- Intersection of \( x + y = 10 \) and \( 3x + 2y = 120 \).- Intersection of \( x + y = 30 \) and \( 3x + 2y = 120 \).- Include points like (10,0) from the \( x \)-axis and (0,10) from the \( y \)-axis if applicable.

Step 6: Calculate Profit at Each Corner Point

Substitute each corner point back into the objective function \( P = 1.20x + 1.80y \) to calculate the profit:For example: 1. If \( x = 20, y = 0 \): \[ P = 1.20(20) + 1.80(0) = 24 \]2. If \( x = 0, y = 30 \): \[ P = 1.20(0) + 1.80(30) = 54 \]Continue for all corner points.

Step 7: Select Maximum Profit

Compare the calculated profit for each corner point and choose the one with the maximum value as the optimum solution.For instance, if the highest profit value is \( P = 54 \), then the point with this value identifies the number of dozens of each type of donut to maximize profit.

Key Concepts

Objective Function

In linear programming, the objective function is a mathematical formula that helps us maximize or minimize some quantity, such as profit or cost. It's like setting a target you want to achieve, given some limitations or conditions that you have to work with.
For the donut store exercise, our goal is to maximize the profit from selling two types of donuts: crème-filled and jelly-filled. We define our objective function as follows:
\( P = 1.20x + 1.80y \)
This formula represents total profit, where \( x \) is the number of dozens of crème-filled donuts and \( y \) is the number of dozens of jelly-filled donuts. The goal is to find values for \( x \) and \( y \) that give us the highest possible value for \( P \).
Understanding the objective function helps in focusing the problem toward what truly matters—in this case, profit maximization.

Constraints

Constraints are the conditions or restrictions placed on a problem to ensure that the solution is realistic and workable. For our donut store, constraints dictate how many donuts we can prepare given available resources and demand.
Here are the constraints for this problem:

• The total number of dozens sold must be between 10 and 30: \(10 \leq x + y \leq 30\)
• The total preparation time cannot exceed 120 minutes, with 3 minutes for each dozen crème-filled and 2 minutes for each dozen jelly-filled: \(3x + 2y \leq 120\)
• The numbers of dozens \(x\) and \(y\) cannot be negative: \(x \geq 0, y \geq 0\)

These constraints are crucial for forming the feasible region, which tells us where on a graph the acceptable solutions could be located.

Feasible Region

The feasible region in a linear programming problem is the area on a graph where all constraints are satisfied. It represents all the possible solutions that meet the problem's requirements.
For the donut problem, we plot each constraint on a graph to visualize their overlap. The overlapping area is the feasible region.
Here's how we identify the feasible region:

• The line \(x + y = 10\) sets a lower limit.
• The line \(x + y = 30\) sets an upper limit.
• The line \(3x + 2y = 120\) angles downward, affecting how the region is shaped.

The feasible region is the shaded area where all conditions are satisfied simultaneously. It contains the points that are potential solutions to our objective function. Identifying this area is vital for determining which values of \(x\) and \(y\) are possible for maximizing profit.

Profit Maximization

Profit maximization is the process of finding the best possible solution that gives the maximum profit within the constraints of the problem. In our donut store scenario, we achieve this by evaluating the objective function \(P = 1.20x + 1.80y\) at the vertices, or corner points, of the feasible region.
The corner points represent potential combinations of dozens of donuts that meet all constraints. By substituting these points into the objective function, we identify which combination yields the highest profit.
For instance, at one corner point \((x = 0, y = 30)\), the profit is \(P = 54\). Finding the highest profit among all evaluated corner points leads to the solution that maximizes profit. This strategic analysis helps bakery managers deliver the best financial outcome given the operational constraints.