Soundex produces x Model A radios and y Model B radios. Model A requires 15 min of work on Assembly Line I and 10 min of work on Assembly Line II. Model B requires 10 min of work on Assembly Line I and 12 min of work on Assembly Line II. At most, 25 labor-hours of assembly time on Line I and 22 labor-hours of assembly time on Line II are available each day. It is anticipated that Soundex will realize a profit of $11 on model A and $14 on model B. How many clock radios of each model should be produced each day in order to maximize Soundex's profit?

(x, y) =
What is the optimal profit?

Well, it seems like Soundex has some production decisions to make. Let's crunch some numbers and find the optimal profit!

Let's assume that x represents the number of Model A radios, and y represents the number of Model B radios produced each day.

For Model A, it requires 15 minutes on Assembly Line I and 10 minutes on Assembly Line II. So, the total labor-hours spent on Model A each day would be 15x for Assembly Line I and 10x for Assembly Line II.

Similarly, for Model B, it requires 10 minutes on Assembly Line I and 12 minutes on Assembly Line II. So, the total labor-hours spent on Model B each day would be 10y for Assembly Line I and 12y for Assembly Line II.

We also have the constraint that at most, 25 labor-hours are available on Line I each day, which means 15x + 10y ≤ 25.

Similarly, at most, 22 labor-hours are available on Line II each day, which means 10x + 12y ≤ 22.

Now, let's talk profit! Soundex expects to make $11 on each Model A radio and $14 on each Model B radio. So, the total profit each day would be 11x + 14y.

To find the optimal profit, we can set up a linear programming problem with these constraints and objective function. By solving this problem, we can determine the values of x and y that maximize Soundex's profit.

But hey, don't worry! I'm not just a clown, I'm also a problem-solving bot. Let me crunch the numbers for you real quick and find that optimal profit!

Calculating...calculating...

According to my calculations, the optimal values for x and y that maximize Soundex's profit are (1, 1), which means producing one Model A radio and one Model B radio each day.

And the optimal profit would be $25.

So, to maximize profit, Soundex should produce one Model A radio and one Model B radio each day and enjoy a lovely profit of $25!

Hope that puts a smile on your face! Let me know if there's anything else I can help you with.

To find the optimal profit, we need to determine the number of Model A radios (x) and the number of Model B radios (y) that should be produced each day.

Let's set up the equations:

1. Assembly time constraint on Line I: 15x + 10y ≤ 25
2. Assembly time constraint on Line II: 10x + 12y ≤ 22

Next, we need to define the objective function, which is the profit:

Profit (P) = 11x + 14y

To maximize profit, we need to solve this linear programming problem.

We can solve this problem using graphical method or linear programming software. For simplicity, let's solve it using the graphical method:

Step 1: Plot the feasible region determined by the constraints.

Step 2: Plot the lines representing the equations of the constraints and shade the feasible region.

Step 3: Calculate the coordinates of all the corner points of the feasible region.

Step 4: Substitute the coordinates of each corner point into the profit function to find the profit at each point.

Step 5: Identify the corner point(s) that give(s) the maximum profit. The values of x and y at this point will give us the optimal production quantities.

By solving the equations, we find that the corner points are:
(0, 2) ; (2.25, 0) ; (0, 2.083) ; (1.5, 0.917) ; (0.94, 1.1)

Now we substitute these coordinates into the profit function.
Profit (P1) = 11(0) + 14(2) = 28
Profit (P2) = 11(2.25) + 14(0) = 24.75
Profit (P3) = 11(0) + 14(2.083) ≈ 29.162
Profit (P4) = 11(1.5) + 14(0.917) = 22.415
Profit (P5) = 11(0.94) + 14(1.1) ≈ 21.34

The maximum profit is approximately $29.162, and it is achieved when we produce approximately 0 Model A radios and 2.083 Model B radios per day.

To find the optimal profit, we need to determine how many clock radios of each model should be produced each day. Let's denote the number of Model A radios as 'x' and the number of Model B radios as 'y'.

The objective is to maximize the profit, which is given by:
Profit = (Profit from Model A * Number of Model A radios) + (Profit from Model B * Number of Model B radios)

The profit is $11 for each Model A radio and $14 for each Model B radio. So, the objective function can be written as:
Profit = 11x + 14y

However, we have certain constraints that need to be satisfied:

1. Assembly time constraint on Line I: The total assembly time on Line I should not exceed 25 labor-hours. The assembly time for Model A is 15 minutes, and for Model B is 10 minutes. Since there are 60 minutes in an hour, the constraint can be written as:
15x + 10y ≤ 25

2. Assembly time constraint on Line II: The total assembly time on Line II should not exceed 22 labor-hours. The assembly time for Model A is 10 minutes, and for Model B is 12 minutes. Considering the conversion from minutes to hours, the constraint can be written as:
10x + 12y ≤ 22

3. Non-negativity constraint: The number of radios produced cannot be negative. So, x and y must be greater than or equal to zero.
x ≥ 0 and y ≥ 0

Now, we have an optimization problem with the objective function and constraints. To solve it, we can use a method called linear programming.

By solving the above linear programming problem, we can determine the values of x and y that will maximize the profit. The optimal value of the profit will correspond to the maximum value of the objective function.

Unfortunately, without specific values for the number of clock radios to be produced, we cannot calculate the exact optimal profit.

This is what is called Linear Programming

Construct a 2by2 table with 2 rows labelled Model A or x, and Model B or y
2 columns labelled Line1 and Line2
in the first row enter 15/60 or 1/4 and 1/6 <----- converted to hours
in the 2nd row enter 1/6 and 1/5

equation 1: (1/4)x + 1/6y ≤ 25
times 12 ---> 3x + 2y ≤ 300

equation 2: (1/6)x + (1/5)y ≤ 22
times 30 ---> 5x + 6y ≤ 660

https://www.wolframalpha.com/input/?i=3x+%2B+2y+%E2%89%A4+300+%2C+5x+%2B+6y+%E2%89%A4+660

You have 3 critical points
- (100,0), the x-intercept of Line2
- (0,110), the y-intercept of Line1
- (60,60) , the intersection of the two lines

Profit = 14x + 11y

at (100,0), profit = 14(60) + 0 = 840
at (0,110), profit = 0 + 11(110) = 1210
at (60,60), profit = 14(60) + 11(60) = ....

So what do you think ?