Math Problem Statement

ITEM 5 A fashion designer makes two types of designs; one design on trousers and another design on dresses. He takes 3 hours to make a design on a trouser and 4 hours to make a design on a dress. He works for a maximum of 120 hours to make designs on trouser and dresses. It costs him shs 4000 to make a design on a trouser and
shs 1500 to make a design on a dress. The total cost does not exceed
shs 90,000. He must make designs on trousers for at least 8 trousers and make designs on dresses for more than 12 dresses. He makes a profit of shs 400 on each trouser and shs 700 on each dress. Task: a) Write down mathematical statements that shows the relationship between the trousers and dresses. b) Show the feasible region of the relation on the Cartesian plane. c) Help the designer to determine the maximum possible profit he makes.

Solution

Task a) Write down mathematical statements that show the relationship between trousers and dresses.

Let:

  • xx be the number of trousers.
  • yy be the number of dresses.

Constraints:

  1. Time constraint:
    The designer works for a maximum of 120 hours. It takes 3 hours for a trouser design and 4 hours for a dress design. Thus: 3x+4y1203x + 4y \leq 120

  2. Cost constraint:
    The designer spends shs 4000 on each trouser design and shs 1500 on each dress design. The total cost should not exceed shs 90,000. Therefore: 4000x+1500y90,0004000x + 1500y \leq 90,000

  3. Minimum number of trousers:
    He must make at least 8 trousers: x8x \geq 8

  4. Minimum number of dresses:
    He must make more than 12 dresses: y>12y > 12

  5. Non-negativity constraint:
    Both xx and yy must be non-negative because the number of trousers and dresses cannot be negative: x0andy0x \geq 0 \quad \text{and} \quad y \geq 0

Profit:

The profit for each trouser is shs 400, and for each dress is shs 700. Therefore, the total profit function is: Profit=400x+700y\text{Profit} = 400x + 700y


Task b) Show the feasible region of the relation on the Cartesian plane.

To show the feasible region on the Cartesian plane, we need to plot the following inequalities:

  1. 3x+4y1203x + 4y \leq 120
  2. 4000x+1500y90,0004000x + 1500y \leq 90,000
  3. x8x \geq 8
  4. y>12y > 12
  5. x0x \geq 0, y0y \geq 0

Steps for plotting:

  1. Plot the line 3x+4y=1203x + 4y = 120:

    • Find the intercepts:
      • xx-intercept: 3x=120x=403x = 120 \Rightarrow x = 40.
      • yy-intercept: 4y=120y=304y = 120 \Rightarrow y = 30.
    • Draw the line through these points and shade below it (since it's \leq).
  2. Plot the line 4000x+1500y=90,0004000x + 1500y = 90,000:

    • Simplify the equation:
      8x+3y=1808x + 3y = 180.
    • Find the intercepts:
      • xx-intercept: 8x=180x=22.58x = 180 \Rightarrow x = 22.5.
      • yy-intercept: 3y=180y=603y = 180 \Rightarrow y = 60.
    • Draw the line through these points and shade below it (since it's \leq).
  3. Plot x8x \geq 8:

    • Draw a vertical line at x=8x = 8 and shade to the right.
  4. Plot y>12y > 12:

    • Draw a horizontal line at y=12y = 12 and shade above it.
  5. Plot x0x \geq 0 and y0y \geq 0:

    • These constraints restrict the feasible region to the first quadrant (i.e., where both x0x \geq 0 and y0y \geq 0).

The feasible region is the intersection of all the shaded areas, and the maximum possible profit occurs at a corner point within this region.


Task c) Help the designer to determine the maximum possible profit.

To maximize the profit 400x+700y400x + 700y, we need to evaluate the profit at the corner points of the feasible region found in part b. These points are the solutions to the system of inequalities.

Steps:

  1. Find the corner points by solving the system of equations corresponding to the boundary lines of the feasible region.

    • Solve for the intersections of:
      • 3x+4y=1203x + 4y = 120 and 8x+3y=1808x + 3y = 180 (intersection of time and cost constraints).
      • 3x+4y=1203x + 4y = 120 and x=8x = 8 (intersection of time constraint and minimum trousers).
      • 8x+3y=1808x + 3y = 180 and y=12y = 12 (intersection of cost constraint and minimum dresses).
  2. Substitute the coordinates of the corner points into the profit function 400x+700y400x + 700y to determine the maximum profit.

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Math Problem Analysis

Mathematical Concepts

Linear Inequalities
Linear Programming
Profit Maximization

Formulas

3x + 4y ≤ 120
4000x + 1500y ≤ 90000
x ≥ 8
y > 12
Profit = 400x + 700y

Theorems

Linear Programming Optimization

Suitable Grade Level

Grades 10-12