To solve the problem of determining the feasible region for the number of volleyballs (x) and soccer balls (y) the coach can order, we will need to set up inequalities based on the information provided.
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Cost Constraints:
- The total cost of the volleyballs can be represented as \(20x\) (since each volleyball costs $20).
- The total cost of the soccer balls can be represented as \(25y\) (since each soccer ball costs $25).
- The coach can spend a maximum of $1,100, leading to the inequality: \[ 20x + 25y \leq 1100 \]
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Quantity Constraints:
- The coach plans to order at least 50 volleyballs and soccer balls in total. This gives us the inequality: \[ x + y \geq 50 \]
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Non-negativity Constraints:
- Since the number of volleyballs and soccer balls cannot be negative, we also have: \[ x \geq 0 \quad \text{and} \quad y \geq 0 \]
Graphing the Inequalities
Let's graph these inequalities to find the feasible region.
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Graphing \(20x + 25y \leq 1100\):
- To find the intercepts, set \(y = 0\): \[ 20x = 1100 \quad \Rightarrow \quad x = 55 \quad \text{(x-intercept)} \]
- Set \(x = 0\): \[ 25y = 1100 \quad \Rightarrow \quad y = 44 \quad \text{(y-intercept)} \]
- The line \(20x + 25y = 1100\) will have x-intercept (55, 0) and y-intercept (0, 44).
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Graphing \(x + y \geq 50\):
- To find the intercepts, set \(y = 0\): \[ x = 50 \quad \text{(x-intercept)} \]
- Set \(x = 0\): \[ y = 50 \quad \text{(y-intercept)} \]
- The line \(x + y = 50\) will have x-intercept (50, 0) and y-intercept (0, 50).
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Non-negativity Constraints:
- The feasible region will be in the first quadrant where \(x \geq 0\) and \(y \geq 0\).
Finding the Feasible Region
- The region satisfying both inequalities lies below the line \(20x + 25y = 1100\) and above or on the line \(x + y = 50\).
- The intersections of these lines will also need to be calculated to determine corner points that make up the feasible region.
Finding Intersection Points
- Set the equations equal to find the intersection:
- From the equations: \[ 20x + 25y = 1100 \] \[ x + y = 50 \quad \Rightarrow \quad y = 50 - x \]
- Substitute \(y\) in the first equation: \[ 20x + 25(50 - x) = 1100 \] \[ 20x + 1250 - 25x = 1100 \] \[ -5x + 1250 = 1100 \quad \Rightarrow \quad -5x = -150 \quad \Rightarrow \quad x = 30 \]
- Substitute \(x = 30\) back into \(y = 50 - x\): \[ y = 50 - 30 = 20 \]
- So, the intersection point is \((30, 20)\).
Summary
Thus, the feasible region is defined by the points where:
- The line \(20x + 25y = 1100\) intersects the axes.
- The line \(x + y = 50\) intersects the axes.
- And other points in the first quadrant.
You can now mark the points (0, 44), (55, 0), (50, 0), (0, 50), and (30, 20) on a graph, and the area that satisfies all inequalities is the feasible region where the coach can choose how many volleyballs and soccer balls to order within budget and quantity limits.
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