Ok im stumped on this HW problem can anyone help?
The Blue clothing company makes both shirts and pants. Each shirt gives a profit of $25, while a pair of pants yields $55. Both shirts and pants are made using machines and human labor. A single shirt takes an hour of machine labor and an hour of human labor. A pair of pants requires 3 hours of machine labor and 2 hours of human labor. Each day the company can devote a total of 160 hours of human labor and 200 of machine labor. The company must make at least 15 pairs of pants every day, but never has enough material to make more than 70. The number of shirts made in a day can not exceed 100. Lastly, the number of pants made in a day can never be more than the number of shirts. How many pants and shirts should the company make in a day to maximize profits? What is the maximum profit?
So far i set up the profit equation 25x+55y=Profit and now im stuck on the different restrictions for it i think i have y≤70 x≤100 plz help!
human hrs = 160, or 2x+6y=160
machine hrs = 200, or 2x+9y=200
minimum of 15 pants, or y>=15
must make more shirts than pants, or
x>=y
the maximum profit is where the equation 25x+55y=profit has a local maximum. One way to determine this is using the 1st derivative and set this equal to zero.
hey thanks alot for all that help i graphed it on some paper and in my calculator and i think for the answer i got 18.18 for x and 18.18 for y and the profit would be $1454.54 can anyone confirm if this is correct? thanks
x=shirt y=pants
i got 18.18 for x and 18.18 for y and the profit would be $1454.54 does that look like its right?
I got a maximum profit of $4200 with 40 pants and 80 shirts.
To solve this problem, we need to consider the given constraints and find the values of x and y that maximize the profit.
Let's begin by creating a list of the given constraints:
1. The total available machine labor is limited to 200 hours per day.
=> Machine labor for shirts: x hours
=> Machine labor for pants: 3y hours
So, the constraint for machine labor is: x + 3y ≤ 200
2. The total available human labor is limited to 160 hours per day.
=> Human labor for shirts: x hours
=> Human labor for pants: 2y hours
So, the constraint for human labor is: x + 2y ≤ 160
3. The company must make at least 15 pairs of pants every day.
=> Constraint for pants: y ≥ 15
4. The company cannot make more than 70 pants due to material limitations.
=> Constraint for pants: y ≤ 70
5. The number of shirts made in a day cannot exceed 100.
=> Constraint for shirts: x ≤ 100
6. The number of pants made in a day cannot be more than the number of shirts.
=> Constraint for pants and shirts: y ≤ x
Now, we have a system of constraints:
x + 3y ≤ 200
x + 2y ≤ 160
y ≥ 15
y ≤ 70
x ≤ 100
y ≤ x
To solve this system, we can use graphical methods or linear programming techniques. Here, we will use graphical methods.
First, plot the feasible region by graphing the inequalities on a graph, keeping in mind the given limitations:
The feasible region:
- The shaded region where all the constraints are satisfied.
Next, find the corner points of the feasible region. These corner points are the intersection points of the lines that represent the constraints.
The corner points are:
A: (0, 70)
B: (0, 15)
C: (85, 15)
D: (100, 100)
E: (85, 57.5)
Now, we need to find the profit at each corner point to determine which point maximizes the profit.
Using the profit equation: Profit = 25x + 55y
Profit at point A: Profit_A = 25(0) + 55(70) = $3,850
Profit at point B: Profit_B = 25(0) + 55(15) = $825
Profit at point C: Profit_C = 25(85) + 55(15) = $2,375
Profit at point D: Profit_D = 25(100) + 55(100) = $8,250
Profit at point E: Profit_E = 25(85) + 55(57.5) = $5,162.50
From the calculations, we can conclude that the maximum profit is achieved at point D: (100, 100), with a profit of $8,250.
Therefore, the company should make 100 shirts and 100 pants to maximize profits, resulting in a maximum profit of $8,250.