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:
- be the number of trousers.
- be the number of dresses.
Constraints:
-
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: -
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: -
Minimum number of trousers:
He must make at least 8 trousers: -
Minimum number of dresses:
He must make more than 12 dresses: -
Non-negativity constraint:
Both and must be non-negative because the number of trousers and dresses cannot be negative:
Profit:
The profit for each trouser is shs 400, and for each dress is shs 700. Therefore, the total profit function is:
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:
- ,
Steps for plotting:
-
Plot the line :
- Find the intercepts:
- -intercept: .
- -intercept: .
- Draw the line through these points and shade below it (since it's ).
- Find the intercepts:
-
Plot the line :
- Simplify the equation:
. - Find the intercepts:
- -intercept: .
- -intercept: .
- Draw the line through these points and shade below it (since it's ).
- Simplify the equation:
-
Plot :
- Draw a vertical line at and shade to the right.
-
Plot :
- Draw a horizontal line at and shade above it.
-
Plot and :
- These constraints restrict the feasible region to the first quadrant (i.e., where both and ).
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 , 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:
-
Find the corner points by solving the system of equations corresponding to the boundary lines of the feasible region.
- Solve for the intersections of:
- and (intersection of time and cost constraints).
- and (intersection of time constraint and minimum trousers).
- and (intersection of cost constraint and minimum dresses).
- Solve for the intersections of:
-
Substitute the coordinates of the corner points into the profit function 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
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