Question 1189926: Given:
Room1 : length=4m : width=3m
Room2 : length=6m : width=4m
Tiles:
Glossy1 : 0.5m x 0.5m cost $8/tile
Glossy2 : 1m x 1m cost $10/tile
Non glossy1: 0.5m x 0.5m cost $5/tile
Non glossy2: 1m x 1m cost $7/tile
How to formulate 2 quadratic inequalities involving the dimensions of the floor of the rooms and the measure and costs of the tiles. Then graph the solution sets of these inequalities.
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to formulate the quadratic inequalities and approach the graphing:
**1. Area Calculations:**
* **Room 1 Area:** Length * Width = 4m * 3m = 12 m²
* **Room 2 Area:** Length * Width = 6m * 4m = 24 m²
**2. Tile Area Calculations:**
* **Glossy 1 Tile Area:** 0.5m * 0.5m = 0.25 m²
* **Glossy 2 Tile Area:** 1m * 1m = 1 m²
* **Non-Glossy 1 Tile Area:** 0.5m * 0.5m = 0.25 m²
* **Non-Glossy 2 Tile Area:** 1m * 1m = 1 m²
**3. Number of Tiles Needed (Variables):**
Let's define variables for the number of each type of tile:
* g1 = Number of Glossy 1 tiles
* g2 = Number of Glossy 2 tiles
* n1 = Number of Non-Glossy 1 tiles
* n2 = Number of Non-Glossy 2 tiles
**4. Area Inequalities:**
Since the total tile area must be *at least* equal to the room area, we have two inequalities for each room:
* **Room 1:**
* 0.25g1 + 1g2 + 0.25n1 + 1n2 ≥ 12
* **Room 2:**
* 0.25g1 + 1g2 + 0.25n1 + 1n2 ≥ 24
**5. Cost Inequalities:**
Let's assume a combined budget for both rooms. Let 'B' be the total budget. The cost inequalities are:
* 8g1 + 10g2 + 5n1 + 7n2 ≤ B
**6. Graphing the Solution Sets:**
Graphing these inequalities directly in 4D space (g1, g2, n1, n2) is impossible on a 2D surface. To visualize, we need to make some simplifications. Here are a few approaches:
* **Scenario 1: Fixed Tile Mix:** Assume a fixed ratio of tile types (e.g., equal numbers of glossy and non-glossy tiles). This reduces the variables and makes graphing possible. For example, if g1 = n1 and g2 = n2, our inequalities would be:
* Room 1: 0.5g1 + 2g2 ≥ 12
* Room 2: 0.5g1 + 2g2 ≥ 24
* Cost: 13g1 + 17g2 ≤ B
Now you have inequalities in 2D space (g1 and g2) that you can graph.
* **Scenario 2: Focus on One Room, Two Tile Types:** Consider just Room 1 and two tile types (e.g., Glossy 1 and Glossy 2). The inequalities become:
* 0.25g1 + g2 ≥ 12
* 8g1 + 10g2 ≤ B (budget for Room 1)
Again, this is a 2D graphing problem (g1 and g2).
* **Software:** Software like GeoGebra or Wolfram Alpha can handle 3D graphing, which could be used if you can reduce the number of variables to three.
**7. Interpreting the Graphs:**
The solution set (the feasible region) on the graph represents the combinations of tile numbers that satisfy both the area and cost constraints. Any point within this region represents a possible solution.
**Example (Scenario 2):**
Let's say the budget for Room 1 is $200. Our inequalities are:
* 0.25g1 + g2 ≥ 12
* 8g1 + 10g2 ≤ 200
You would graph these two inequalities on a g1-g2 plane. The overlapping shaded region is your solution set.
**Important Considerations:**
* **Integer Solutions:** Since you can't buy fractions of tiles, you're only interested in integer solutions within the feasible region.
* **Optimization:** You could add an objective function (e.g., minimize cost) and use linear programming techniques to find the optimal solution.
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
Given:
Room1 : length=4m : width=3m
Room2 : length=6m : width=4m
Tiles:
Glossy1 : 0.5m x 0.5m cost $8/tile
Glossy2 : 1m x 1m cost $10/tile
Non glossy1: 0.5m x 0.5m cost $5/tile
Non glossy2: 1m x 1m cost $7/tile
How to formulate 2 quadratic inequalities involving the dimensions of the floor of the rooms
and the measure and costs of the tiles. Then graph the solution sets of these inequalities.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Unfortunately, this problem does not state, what is the goal of this exercise.
I will suppose that the goal is to find the most cheaper cover of the floor in two rooms.
Comparing the cost of 4 Glossy1 (0.5m x 0.5m at $8) against 1 Glossy2 (1m x 1m at $10),
we see that 1 Glossy2 is cheaper for one (and for each) square meter.
Comparing the cost of 4 non-glossy1 (0.5m x 0.5m at $5) against 1 non-glossy2 (1m x 1m at $7),
we see that 1 non-glossy2 is cheaper for one (and for each) square meter.
Comparing the cost of 1 Glossy2 (1m x 1m at $10) against 1 non-glossy2 (1m x 1m at $7),
we see that 1 non-glossy2 is cheaper for one (and for each) square meter.
So, the cheapest option for all cases and from the point of view of different combinations,
is the choice of non-glossy2 (1m x 1m at $7).
It gives the total cost for the total area of the two rooms (4*3 + 6*4)*7 = 252 dollars.
ANSWER. The cheapest cover for the two room is non-glossy2 (1m x 1m at $7),
giving the total cost of 252 dollars.
Solved mentally.
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In my view, my interpretation, my approach and my solution transform this
(originally non-sensical) problem in something that makes sense.
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