A small shoe company makes two styles of shoes, Oxford (x) and Loafers (y). The following constraints are given: x>=30, 10<=y<=70, x+y<=120. The profit made on the sale of the shoes is given by P=315x + 280y. Use the boundary-search method to calculate the maximum profit value.

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Asked on July 30, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-07-30T20:43:10.493Z

Step 1: Identify the constraints and vertices of the feasible region (for 4.1). The given constraints are: 1. x 30 2. 10 y 70 (which means y 10 and y 70) 3. x + y 120 To sketch the feasible region, we plot the boundary lines and determine the region that satisfies all inequalities. The vertices of this feasible region are the intersection points of these boundary lines: Line 1: x = 30 Line 2: y = 10 Line 3: y = 70 Line 4: x + y = 120 Let's find the vertices: Vertex A: Intersection of x=30 and y=10. A = (30, 10) Vertex B: Intersection of x=30 and y=70. B = (30, 70) Vertex C: Intersection of y=70 and x+y=120. Substitute y=70 into x+y=120 x+70=120 x=50. C = (50, 70) Vertex D: Intersection of y=10 and x+y=120. Substitute y=10 into x+y=120 x+10=120 x=110. D = (110, 10) For part 4.1, you would plot these lines on a graph and shade the polygon formed by these vertices. The feasible region is the area bounded by the points (30,10), (30,70), (50,70), and (110,10). Step 2: Calculate the profit at each vertex (for 4.2). The profit function is P = 315x + 280y. We evaluate this function at each vertex of the feasible region: At A=(30, 10): P_A = 315(30) + 280(10) = 9450 + 2800 = 12250 At B=(30, 70): P_B = 315(30) + 280(70) = 9450 + 19600 = 29050 At C=(50, 70): P_C = 315(50) + 280(70) = 15750 + 19600 = 35350 At D=(110, 10): P_D = 315(110) + 280(10) = 34650 + 2800 = 37450 The maximum profit value is the largest of these values. The maximum profit value is 37450. Step 3: Determine the number of shoes for maximum profit (for 4.3). The maximum profit of 37450 occurs at vertex D=(110, 10). This means the company must sell x=110 Oxfords and y=10 Loafers to achieve maximum profit. The company must sell 110 Oxfords and 10 Loafers. Step 4: Check feasibility for 60 Oxfords and 50 Loafers (for 4.4). We need to check if the point (x, y) = (60, 50) satisfies all the given constraints: 1. x 30 60 30 (True) 2. y 10 50 10 (True) 3. y 70 50 70 (True) 4. x + y 120 60 + 50 120 110 120 (True) Since all constraints are satisfied, it is feasible for the company to make 60 Oxfords and 50 Loafers. The answer is Yes, it is feasible. Reason: The point (60, 50) satisfies all the given inequalities (x 30, 10 y 70, and x+y 120), meaning it lies within the feasible region. What's next?

Accepted Answer · 2026-07-30T20:43:10.493Z

Step 1: Identify the constraints and vertices of the feasible region (for 4.1). The given constraints are: 1. x 30 2. 10 y 70 (which means y 10 and y 70) 3. x + y 120 To sketch the feasible region, we plot the boundary lines and determine the region that satisfies all inequalities. The vertices of this feasible region are the intersection points of these boundary lines: Line 1: x = 30 Line 2: y = 10 Line 3: y = 70 Line 4: x + y = 120 Let's find the vertices: Vertex A: Intersection of x=30 and y=10. A = (30, 10) Vertex B: Intersection of x=30 and y=70. B = (30, 70) Vertex C: Intersection of y=70 and x+y=120. Substitute y=70 into x+y=120 x+70=120 x=50. C = (50, 70) Vertex D: Intersection of y=10 and x+y=120. Substitute y=10 into x+y=120 x+10=120 x=110. D = (110, 10) For part 4.1, you would plot these lines on a graph and shade the polygon formed by these vertices. The feasible region is the area bounded by the points (30,10), (30,70), (50,70), and (110,10). Step 2: Calculate the profit at each vertex (for 4.2). The profit function is P = 315x + 280y. We evaluate this function at each vertex of the feasible region: At A=(30, 10): P_A = 315(30) + 280(10) = 9450 + 2800 = 12250 At B=(30, 70): P_B = 315(30) + 280(70) = 9450 + 19600 = 29050 At C=(50, 70): P_C = 315(50) + 280(70) = 15750 + 19600 = 35350 At D=(110, 10): P_D = 315(110) + 280(10) = 34650 + 2800 = 37450 The maximum profit value is the largest of these values. The maximum profit value is 37450. Step 3: Determine the number of shoes for maximum profit (for 4.3). The maximum profit of 37450 occurs at vertex D=(110, 10). This means the company must sell x=110 Oxfords and y=10 Loafers to achieve maximum profit. The company must sell 110 Oxfords and 10 Loafers. Step 4: Check feasibility for 60 Oxfords and 50 Loafers (for 4.4). We need to check if the point (x, y) = (60, 50) satisfies all the given constraints: 1. x 30 60 30 (True) 2. y 10 50 10 (True) 3. y 70 50 70 (True) 4. x + y 120 60 + 50 120 110 120 (True) Since all constraints are satisfied, it is feasible for the company to make 60 Oxfords and 50 Loafers. The answer is Yes, it is feasible. Reason: The point (60, 50) satisfies all the given inequalities (x 30, 10 y 70, and x+y 120), meaning it lies within the feasible region. What's next?

A small shoe company makes two styles of shoes, Oxford (x) and Loafers (y). The following constraints are given: x>=30, 10<=y<=70, x+y<=120. The profit made on the sale of the shoes is given by P=315x + 280y. Use the boundary-search method to calculate the maximum profit value.

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A small shoe company makes two styles of shoes, Oxford (x) and Loafers (y). The following constraints are given: x>=30, 10<=y<=70, x+y<=120. The profit made on the sale of the shoes is given by P=315x + 280y. Use the boundary-search method to calculate the maximum profit value.

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