Bakery has bought 250 pounds of muffin dough. They want to make waffles or muffins in half-dozen packs out of it. Half a dozen of muffins requires 1 lb of dough and a pack of waffles uses 3/4 lb of dough. It take bakers 6 minutes to make a half-dozen of waffles and 3 minutes to make a half-dozen of muffins. Their profit will be $1.50 on each pack of waffles and $2.00 on each pack of muffins. How many of each should they make to maximize profit, if they have just 20 hours to do everything?
To maximize profit, we need to determine the number of packs of waffles and muffins to make. Let's assume the number of packs of waffles is represented by 'w' and the number of packs of muffins is represented by 'm'.
We are given the following information:
- Half a dozen of muffins requires 1 lb of dough.
- A pack of waffles uses 3/4 lb of dough.
- It takes 6 minutes to make a half-dozen of waffles.
- It takes 3 minutes to make a half-dozen of muffins.
- The profit per pack of waffles is $1.50.
- The profit per pack of muffins is $2.00.
- There are 250 pounds of muffin dough available.
- There are 20 hours available in total.
To solve this problem, we need to consider two constraints:
1) The weight of dough used should not exceed the available 250 pounds.
2) The time taken to make the products should not exceed the available 20 hours.
Let's set up the equations to represent the problem:
Equation 1: Weight of dough used for muffins: 1 * m
Equation 2: Weight of dough used for waffles: (3/4) * w
Equation 3: Total dough weight constraint: (1 * m) + ((3/4) * w) <= 250
Equation 4: Time taken for making muffins: 3 * m
Equation 5: Time taken for making waffles: 6 * w
Equation 6: Total time constraint: (3 * m) + (6 * w) <= 20 * 60 (converting hours to minutes)
Now, we need to maximize the profit. The profit from muffins is $2.00 per pack, and the profit from waffles is $1.50 per pack. So, the objective function for maximizing profit is:
Objective Function: P = (2 * m) + (1.5 * w)
To find the optimal solution, we can solve this linear programming problem using the above equations and constraints.
To solve this problem, we need to figure out the number of packs of waffles and muffins that should be made to maximize profit within the time constraints. We can approach this problem using the concept of linear programming.
Let's assume x represents the number of half-dozen packs of muffins, and y represents the number of half-dozen packs of waffles.
We have the following constraints:
1. The weight of dough used for muffins: x pounds
2. The weight of dough used for waffles: (3/4 * y) pounds
3. The time required to make muffins: 3 minutes per x
4. The time required to make waffles: 6 minutes per y
5. The total time available: 20 hours = 1200 minutes
Now, we can formulate the linear programming problem as follows:
Maximize Profit:
Profit(x, y) = 2x + 1.5y
Subject to Constraints:
1. Dough constraint: x + (3/4 * y) <= 250
2. Time constraint: 3x + 6y <= 1200
To solve this problem and determine the optimal values of x and y, we can use a linear programming solver or graphical method. Let's use the graphical method:
Step 1: Convert the inequalities into equations:
x + (3/4 * y) = 250 (Dough constraint)
3x + 6y = 1200 (Time constraint)
Step 2: Plot the graph of these equations on the x-y plane.
Step 3: Identify the feasible region, i.e., the area where both constraints are satisfied, typically represented by shaded region formed by the intersection of the two lines.
Step 4: Find the corner points of the feasible region (where the lines intersect), as these are the potential solutions.
Step 5: Evaluate the objective function (Profit) at each corner point.
Step 6: Choose the corner point that maximizes the Profit. The corresponding values of x and y will give the optimal number of each item to produce.
Following these steps, you can solve the linear programming problem and determine the optimal quantities of muffins (x) and waffles (y) needed to maximize the profit within the given time constraints.
If there are x packs of waffles and y packs of muffins, you want to
maximize p = 1.50x+2.00y subject to
3/4 x + y <= 250
6x + 3y <= 1200