CALCU-Chetney's RTA Woodcarving Co. manufactures t

Solution by Steps

step 1

Let

x

be the number of xylophones and

y

be the number of yo-yos produced. The objective function to maximize profit

P

is given by:

P = 20x + 17y - 2(x + y)

which simplifies to

P = 18x + 15y

step 2

The constraints based on labor hours are: 1. Carpentry:

2x + y \leq 40

2. Finishing:

2x + 2y \leq 50

3. Minimum xylophones:

x \geq 5

4. Non-negativity:

x \geq 0, y \geq 0

step 3

To find the feasible region, we graph the constraints on a coordinate plane. The intersection points of the constraints will help identify the feasible region

step 4

The extreme points of the feasible region can be found by solving the equations of the constraints. The points are: 1.

(5, 0)

2.

(5, 15)

3.

(20, 0)

4.

(10, 10)

step 5

Evaluating the objective function

P

at each extreme point: 1. At

(5, 0)

:

P = 18(5) + 15(0) = 90

2. At

(5, 15)

:

P = 18(5) + 15(15) = 315

3. At

(20, 0)

:

P = 18(20) + 15(0) = 360

4. At

(10, 10)

:

P = 18(10) + 15(10) = 330

step 6

The optimal solution occurs at

(20, 0)

with a maximum profit of

P = 360

Answer

Chetney should manufacture 20 xylophones and 0 yo-yos for a maximum profit of RM 360.

Key Concept

Linear programming is a method to achieve the best outcome in a mathematical model with linear relationships.

Explanation

The solution involves maximizing profit while adhering to constraints on labor hours and minimum production requirements. The optimal production plan is determined by evaluating the objective function at the extreme points of the feasible region.

solve the question one by one with symbol

To solve the problem step by step, let's start with part (a) and formulate the linear programming model. Part (a): Formulate the Linear Programming Model 1. Variables: - Let

x

be the number of xylophones produced. - Let

y

be the number of yo-yos produced. 2. Objective Function: The profit from selling xylophones and yo-yos can be expressed as:

P = 20x + 17y - 2(x + y)

Simplifying this, we get:

P = 18x + 15y

We want to maximize

P

. 3. Constraints: - Carpentry Hours: Each xylophone requires 2 hours and each yo-yo requires 1 hour. The total carpentry hours available is 40:

2x + y \leq 40

- Finishing Hours: Each xylophone requires 2 hours and each yo-yo requires 2 hours. The total finishing hours available is 50:

2x + 2y \leq 50

- Minimum Production of Xylophones: At least 5 xylophones must be produced:

x \geq 5

- Non-negativity Constraints: The number of toys produced cannot be negative:

x \geq 0, \quad y \geq 0

Summary of the Model Maximize:

P = 18x + 15y

Subject to:

\begin{align*} 2x + y &amp; \leq 40 \quad \text{(Carpentry)} \\ 2x + 2y &amp; \leq 50 \quad \text{(Finishing)} \\ x &amp; \geq 5 \quad \text{(Minimum xylophones)} \\ x &amp; \geq 0, \quad y \geq 0 \quad \text{(Non-negativity)} \end{align*}

Part (b): Find the Feasible Region To find the feasible region, we will graph the constraints: 1. Graph the constraints: - For

2x + y \leq 40

: - When

x = 0

,

y = 40

(point (0, 40)) - When

y = 0

,

x = 20

(point (20, 0)) - For

2x + 2y \leq 50

: - When

x = 0

,

y = 25

(point (0, 25)) - When

y = 0

,

x = 25

(point (25, 0)) - The line

x = 5

is a vertical line at

x = 5

. 2. Plot these lines on a graph and identify the area that satisfies all constraints. Part (c): Find the Extreme Points To find the extreme points of the feasible region, we will solve the equations of the lines where they intersect: 1. Intersection of \( 2x + y = 40 \) and \( 2x + 2y = 50 \): - From

2x + y = 40

, we can express

y = 40 - 2x

. - Substitute into

2x + 2(40 - 2x) = 50

:

2x + 80 - 4x = 50 \implies -2x = -30 \implies x = 15

Then,

y = 40 - 2(15) = 10

. So, one extreme point is

(15, 10)

. 2. Other intersections: -

(5, 30)

from

x = 5

and

2x + y = 40

. -

(5, 20)

from

x = 5

and

2x + 2y = 50

. -

(20, 0)

from

2x + y = 40

and

y = 0

. Summary of Extreme Points -

(5, 30)

-

(5, 20)

-

(15, 10)

-

(20, 0)

Part (d): Optimal Solution Now, we evaluate the objective function

P = 18x + 15y

at each extreme point: 1. At

(5, 30)

:

P = 18(5) + 15(30) = 90 + 450 = 540

2. At

(5, 20)

:

P = 18(5) + 15(20) = 90 + 300 = 390

3. At

(15, 10)

:

P = 18(15) + 15(10) = 270 + 150 = 420

4. At

(20, 0)

:

P = 18(20) + 15(0) = 360 + 0 = 360

Conclusion The optimal solution occurs at the point

(5, 30)

with a maximum profit of RM 540. Summary Table of Extreme Points \begin{tabular}{|l|l|} \hline Extreme Points & Values \\ \hline (5, 30) & 540 \\ (5, 20) & 390 \\ (15, 10) & 420 \\ (20, 0) & 360 \\ \hline \end{tabular} If you have any further questions or need clarification on any part, feel free to ask!

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