A furniture company displays bedroom sets which require 21 square meters of space and living room sets which require 42 square meters of space. The company which has 546 square meters of available space,wants to display at least 6 bedroom sets and at least 5 living room sets.

My question is:If a bedroom set sells for $10,000 and a living room set sells for $18,000,determine the number of bedroom sets and living room sets that must be sold to maximize the amount collected.
I'm not sure what I need to do to set this up.

Think it through.. don't try to find a single magic equation.

You maximize receipts when you sell as many living room sets as you can, while still being able to display and sell the required minimum 6 bedroom sets. Since both take up 21 sq meters and you have 546 sq m available, you can display 546/21 = 26 sets of either kind. So, display the required 6 minimum bedroom sets and use the rest of the space for 20 living room sets. Assuming that they all sell at an equal rate (which is not at all likely in practice), the amount collected after one turnover will be 20*$18,000 + 6*$10,000 = $420,000

I did not realize furnishing a house is that expensive nowadays

Correction:

Second last line should say:

check: if b = 6, l = 10 revenue = $240,000

I misread the problem, not noticing the higher area of the living room sets. Just ignore my answer, if you haven't already, and follow Reiny's method.

To solve this problem, we need to set up and solve a linear programming problem, considering the constraints and the objective function.

Let's define:
- x as the number of bedroom sets to be displayed
- y as the number of living room sets to be displayed

We need to maximize the amount collected, which is given by the objective function:
Revenue = ($10,000 * Number of bedroom sets) + ($18,000 * Number of living room sets)

However, we also need to consider the following constraints:
1. The available space constraint:
21 * Number of bedroom sets + 42 * Number of living room sets ≤ 546
(This restriction ensures that the total space required by the displayed sets does not exceed the available space.)

2. The minimum display requirements:
Number of bedroom sets ≥ 6
Number of living room sets ≥ 5

Now, we have set up the problem with the objective function and constraints. To determine the optimal number of bedroom and living room sets, we can use methods like graphing, substitution, or matrix-based methods such as the Simplex algorithm. In this case, we will use the graphing method.

To graph these constraints, we can first rewrite them in slope-intercept form:
- 21x + 42y ≤ 546 => y ≤ (-21/42)x + (546/42) => y ≤ (-1/2)x + 13
- x ≥ 6
- y ≥ 5

Now, let's plot these constraints on a graph and identify the feasible region:
1. Draw the line y = (-1/2)x + 13.
2. Shade the region below the line (since y is less than or equal to the expression in slope-intercept form).
3. Draw a vertical line at x = 6 (since x must be greater than or equal to 6).
4. Shade the region to the right of the vertical line.
5. Draw a horizontal line at y = 5 (since y must be greater than or equal to 5).
6. Shade the region above the horizontal line.

The feasible region is the shaded region where all constraints are satisfied. Now we need to find the corner points of this region to determine the optimal solution.

To find the corner points, we can calculate the intersection points of the lines that form the boundary of the feasible region. In this case, there are four intersection points: (6, 5), (12, 5), (13, 12), and (6, 17).

Next, substitute these corner points into the objective function to calculate the revenue at each point:
- Revenue at (6, 5): ($10,000 * 6) + ($18,000 * 5) = $180,000
- Revenue at (12, 5): ($10,000 * 12) + ($18,000 * 5) = $240,000
- Revenue at (13, 12): ($10,000 * 13) + ($18,000 * 12) = $366,000
- Revenue at (6, 17): ($10,000 * 6) + ($18,000 * 17) = $372,000

Lastly, compare the revenue at each corner point and select the point with the highest revenue. In this case, the point (6, 17) yields the maximum revenue of $372,000.

Therefore, to maximize the amount collected, the furniture company should display 6 bedroom sets and 17 living room sets.

Let the number of bedroom setups be b

then the number of living room setups is (546-21b)/42

total revenue
= 10000b + 18000(546-21b)/42
= 234000 + 1000b

but b ≥ 6 and livingrooms ≥ 5

so if b=6 , l= 10
and if l=5, b = 16

the total revenue is a linear function represented by a straight line with a positive slope, so the larger value of b will give a larger value of revenue

so when b= 16, l = 5 and revenue = $250,000

check: if b = 16, l = 5 revenue = $240,000

So they should set up 16 bedrooms and 5 living rooms