a clothing manufacturer has 100m of silk and 180 m of wool. to make a suit requires 2m of silk and 3m of wool and to make a dress requires 1m of silk and 2m of wool. if the profit on a suit is 108$ and the profit on a dress is 60$ how many suits and dresses should be made to maximize profit?

Question

a clothing manufacturer has 100m of silk and 180 m of wool. to make a suit requires 2m of silk and 3m of wool and to make a dress requires 1m of silk and 2m of wool. if the profit on a suit is 108$ and the profit on a dress is 60$ how many suits and dresses should be made to maximize profit?

This looks like a problem in 'linear programming'

let s represent the number of suits and d the number of dresses.

then 2s + d <= 100 and 3s + 2d <= 180
a d vs
graph these two inequations with s and d as the axes forming a polygon with the s and d axes.

Now consider the equation Profit = 108s + 60d

Shift this equation away from the origin into the first quadrant, until the line p=108s+60d reaches the farthest vertex from the origin of your polygon.

Your polygon only has one vertex in the interior region of the first quadrant, I think it is s=20 and d=30 for a maximum profit of $3960

Answer 1

To solve this problem, we can use the concept of linear programming. This involves setting up constraints and an objective function to maximize or minimize.

Let's define s as the number of suits and d as the number of dresses.

We have the following constraints:

1) The first constraint is the availability of silk: 2s + d <= 100m
This means that the total amount of silk used for suits and dresses cannot exceed 100m.

2) The second constraint is the availability of wool: 3s + 2d <= 180m
This means that the total amount of wool used for suits and dresses cannot exceed 180m.

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

Profit = 108s + 60d

To graph the constraints, we plot them on a graph with s and d as the axes.

Plotting the first constraint: 2s + d <= 100m
Rearranging this equation to solve for d, we get d <= 100 - 2s.
The line d = 100 - 2s represents this constraint.

Plotting the second constraint: 3s + 2d <= 180m
Rearranging this equation to solve for d, we get d <= (180 - 3s)/2.
The line d = (180 - 3s)/2 represents this constraint.

The feasible region is the area where both lines intersect and fall within the positive s and d region.

Next, we calculate the profit at each corner point of the feasible region:

1) (0, 90): Profit = 108(0) + 60(90) = $5400
2) (50, 0): Profit = 108(50) + 60(0) = $5400
3) (20, 30): Profit = 108(20) + 60(30) = $3960 (maximum profit)
4) (0, 0): Profit = 108(0) + 60(0) = $0

The maximum profit of $3960 occurs at the point (20, 30), which means that the clothing manufacturer should make 20 suits and 30 dresses to maximize their profit.