St Joseph's School Has 1,200 Students Each Currently

1 1 22st Josephs School Has 1200 Students Each Of Whom Currently

1. St. Joseph’s School has 1,200 students, each paying $8,000 annually, with an overall budget of $15 million for the upcoming year. The school receives an additional $4.8 million from the church. The total revenue from tuition at the current rate is $9.6 million (1,200 students × $8,000). To cover the entire budget, the school must determine how much to increase tuition per student to avoid a shortfall.

First, calculate the total revenue needed: the total budget is $15 million. Subtracting the church appropriation of $4.8 million leaves $10.2 million to be covered by tuition. Since the current tuition revenue is $9.6 million, there is a shortfall of $600,000 ($10.2 million - $9.6 million). To eliminate this shortfall, the increase in tuition per student should be computed as:

Additional revenue needed per student = $600,000 / 1,200 students = $500

Thus, the new tuition per student should be:

$8,000 + $500 = $8,500

Therefore, the school must raise tuition by $500 per student, setting the new tuition at $8,500, to meet its $15 million budget without a shortfall.

2. Refer to Problem 1-22. Sensing resistance to the idea of raising tuition from members of St. Joseph’s Church, one of the board members suggested that the 960 children of church members could pay $8,000 as usual. Children of nonmembers would pay more. What would the nonmember tuition per year be if St. Joseph’s wanted to continue to plan for a $15 million budget?

Given that 960 students are church members paying $8,000, the revenue from these students is:

960 × $8,000 = $7,680,000

The remaining students are nonmembers: 1,200 - 960 = 240 students. To meet the total required revenue of $15 million, the sum of revenue from church members and nonmembers must equal the total budget less the church’s appropriation, which remains at $4.8 million, thus:

Revenue needed from nonmembers = $15,000,000 - $4,800,000 - $7,680,000 = $2,520,000

To find the nonmember tuition per student:

Nonmember tuition = $2,520,000 / 240 students = $10,500

Therefore, nonmembers would need to pay $10,500 per year to keep the school's budget balanced, assuming church member tuition remains at $8,000 for their students.

3. How should this freighter be loaded to maximize total revenue?

The owner has two cargo options:

• Cargo A: up to 15 tons, volume per ton = 675 ft³, revenue per ton = $85
• Cargo B: up to 54 tons, volume per ton = 450 ft³, revenue per ton = $79

The ship's two holds have volumes:

• Starboard: 14,000 ft³ volume, weight capacity = 26 tons, engine/bridge weight = 6 tons on starboard side.
• Port: 15,400 ft³ volume, weight capacity = 32 tons, and is typically loaded with 6 tons more cargo to balance weight.

The load must be balanced, with equal weight distribution, but considering the engine and bridge on starboard (weighing 6 tons), and the typical loading pattern that results in the port side carrying 6 tons more. The total weight on each side should be equal, accounting for these constraints.

To maximize revenue, the owner should evaluate load options by calculating where to allocate cargo A and B to maximize revenue under volume and weight constraints, along with balancing requirements. The optimal solution involves choosing the combination of cargoes A and B in each hold such that:

• The volume constraints are not exceeded.
• The weight capacities are honored.
• The weights are balanced as per the given constraints.

The ideal approach is to prioritize higher revenue per ton—cargo A at $85 per ton—subject to volume and weight constraints. Given the volume per ton and total volumes, the owner should load as much cargo A as possible into the ship, then fill remaining capacity with cargo B, ensuring that the total weight on each side remains balanced. Exact numeric optimization would involve setting variables for the quantity of each cargo in each hold and solving for maximum revenue via linear programming techniques.

4. How should the two bases be blended to manufacture the two paints at a minimum cost? What is the cost per gallon for each paint?

The problem involves blending bases A and B to produce two types of paint: Tuffcoat and Satinwear, each with specific ingredient constraints and demand levels.

Ingredient proportions and costs are as follows:

The specifications for paint types are:

• Tuffcoat: at least 33% X, at least 35% Y, no more than 14% Z, demand = 1,600 gallons.
• Satinwear: at least 30% X, at least 38% Y, no more than 13% Z, demand = 1,250 gallons.

The objective is to minimize cost by optimally blending bases A and B for each paint, while satisfying ingredient restrictions and demand quantities. This involves setting up linear programming models where variables represent the proportion of each base in each paint, constrained by ingredient content requirements, volume demands, and cost minimization.

The solution process involves calculating the linear combinations that meet the ingredient proportion constraints at minimum cost, which can be achieved through simplex or other LP solving techniques. Based on the proportions and costs, the optimal blends typically favor the base with the lower cost per gallon, adjusted by the ingredient constraints.

For Tuffcoat, blending more of base A is cost-effective given its lower cost unless ingredient constraints demand more of base B. For Satinwear, the same logic applies. The exact minimum costs per gallon for each paint are found by solving the LP models.

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