A toy manufacturer is introducing two new dolls, My First Baby and My Real Baby. In one hour, the company can produce 8 First Babies or 20 Real Babies. Because of demand, the company produces at least twice as many First Babies as Real Babies. The company spends no more than 48 hours per week making these two new dolls. The profit on each First Baby is 3.00$, and the profit on each Real Baby is 7.50$. Find the number and type of dolls that should be produced to maximize profit.
F=2R
R/20 + F/8 < 48
R/20 + 2R/8 < 48
R < 160
F < 320
Well, it sounds like the toy manufacturer has quite the baby dilemma on their hands! Let's see if we can help them figure out the optimal production strategy.
Let's assume that the number of Real Babies produced is x, and since the company produces at least twice as many First Babies as Real Babies, the number of First Babies produced would be 2x.
Now, let's look at the time constraint. In one hour, the company can produce 8 First Babies or 20 Real Babies. So, the number of hours spent on First Babies would be (2x/8), and the number of hours spent on Real Babies would be (x/20). Since there are 48 hours in a week, we get the equation:
(2x/8) + (x/20) ≤ 48.
Simplifying this equation, we get:
10x + 2x ≤ 960.
Combining like terms, we have:
12x ≤ 960.
Dividing both sides by 12, the equation becomes:
x ≤ 80.
So, the maximum number of Real Babies that can be produced is 80.
Now, since the company produces at least twice as many First Babies as Real Babies, the maximum number of First Babies produced would be 2 * 80 = 160.
Finally, let's calculate the profit. The profit on each First Baby is $3.00, so the profit from First Babies would be 3 * 160 = $480.00.
Similarly, the profit on each Real Baby is $7.50, so the profit from Real Babies would be 7.50 * 80 = $600.00.
To maximize profit, the company should produce 160 First Babies and 80 Real Babies.
And there you have it! The optimal production strategy that would yield the highest profit for the toy manufacturer. I hope this helps lighten up the doll dilemma!
To maximize profit, we need to determine the number and type of dolls that should be produced. Let's assume the number of First Babies produced is x, and the number of Real Babies produced is y.
From the given information, we have the following constraints:
1. The company can produce 8 First Babies or 20 Real Babies in 1 hour:
8x + 20y ≤ 48 (since there are 48 hours in a week)
2. The company produces at least twice as many First Babies as Real Babies:
x ≥ 2y
To maximize profit, we need to maximize the objective function, which is the total profit:
Profit = (Profit per First Baby * Number of First Babies) + (Profit per Real Baby * Number of Real Babies)
Profit = (3x) + (7.5y)
With these constraints, we can solve this problem as a linear programming problem.
Step 1: Graph the feasible region by plotting the constraints on a graph:
Let's plot the constraint 8x + 20y ≤ 48
Drawing the graph, we get:
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(x,y)
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Step 2: Determine the corner points of the feasible region:
The feasible region is the area on or below the line 8x + 20y = 48. However, we also have the constraint x ≥ 2y, which narrows down the feasible region. By substituting x = 2y into the equation 8x + 20y = 48, we get 8(2y) + 20y = 48, simplifying to 36y = 48, which gives y = 4/3.
Therefore, the corner points of the feasible region are (8, 4/3) and (0, 0).
Step 3: Calculate the total profit at each corner point:
For (8, 4/3):
Profit = (3 * 8) + (7.5 * 4/3)
Profit ≈ 24 + 10
Profit ≈ 34
For (0, 0):
Profit = (3 * 0) + (7.5 * 0)
Profit = 0
Step 4: Compare the total profits at each corner point:
The maximum profit is obtained at (8, 4/3), with a profit of approximately $34.
Therefore, to maximize profit, the company should produce approximately 8 My First Babies and 4 My Real Babies.
To solve this problem, we can use the concept of linear programming. Linear programming is a mathematical technique used to find the maximum or minimum value of a linear objective function, given a set of linear inequality constraints.
Let's define our variables:
Let x be the number of My First Baby dolls produced.
Let y be the number of My Real Baby dolls produced.
We are trying to maximize the profit, so our objective function is:
Profit = 3x + 7.50y
Now let's write down the constraints:
1. The company can produce 8 First Babies or 20 Real Babies in one hour:
x/8 ≤ y/20
This ensures that the number of First Babies produced is at least twice the number of Real Babies.
2. The company spends no more than 48 hours per week making these two new dolls:
x/8 + y/20 ≤ 48
This accounts for the time constraint.
3. The number of dolls produced cannot be negative:
x ≥ 0 and y ≥ 0
Now we have the objective function and the constraints. To solve this linear programming problem, we can use the Simplex algorithm or any other optimization technique. However, I'll use brute force by checking each possible combination of x and y that satisfies all the constraints, and then calculate the profit for each combination.
Since the constraints involve fractions, we can start by considering integer values for x and y. We can set up a table to calculate the profit for each combination:
| x | y | Profit |
|---|---|--------|
| 0 | 0 | 0 |
| 1 | 0 | 3 |
| 2 | 0 | 6 |
| 3 | 0 | 9 |
|...|...| ...... |
| 8 | 0 | 24 |
| 8 | 1 | 31.5 |
| 8 | 2 | 39 |
|...|...| ...... |
| 8 | 20| 183 |
| 9 | 0 | 27 |
|...|...| ...... |
| 14| 34| 288 |
|...|...| ...... |
We can see that as x increases, the profit increases (due to the higher profit per unit on My First Baby dolls). So to maximize profit, we should produce as many My First Baby dolls as possible while still satisfying the constraints.
From the table, we can see that the maximum profit achievable is $288 when producing 14 My First Baby dolls and 34 My Real Baby dolls.
Therefore, to maximize profit, the company should produce 14 My First Baby dolls and 34 My Real Baby dolls.