Program Linier 01

The total area of the parking lot is 1,760 m2. The average area for a small car is 4 m2 and for a large car is 20 m2. The maximum capacity of the parking lot is 200 vehicles. The cost for parking a small car is Rp 1,000.00 per hour and for a large car is Rp 2,000.00 per hour. If the parking lot is filled to its maximum capacity for one hour without any vehicles leaving or entering, the maximum revenue generated would be calculated by multiplying the maximum capacity (200 vehicles) by the total cost per hour (Rp 1,000.00 for small cars and Rp 2,000.00 for large cars), which equals Rp 260,000.00.

The given problem states that the aircraft can accommodate no more than 48 passengers. Each passenger in the first class can bring 60 kg of luggage, while each passenger in the economy class can bring 20 kg of luggage. The total luggage capacity of the aircraft is 1,440 kg.

To represent this problem mathematically, let x represent the number of passengers in the first class and y represent the number of passengers in the economy class.

Since the total number of passengers cannot exceed 48, the inequality x + y
The total luggage capacity can be represented as 60x + 20y ≤ 1,440. Simplifying this inequality, we get 3x + y
Additionally, since the number of passengers cannot be negative, x > 0 and y > 0.

Therefore, the correct model for this problem is x + y 0; y > 0, which matches option c.

The given model represents a system of inequalities. The objective is to find the maximum value of z = 5x + 3y that satisfies all the given conditions. The feasible region is the area where all the inequalities are true. By graphing the inequalities, it can be observed that the feasible region is bounded by the lines 2x + 3y = 80, 5x + y = 70, x = 0, and y = 0. The maximum value of z = 5x + 3y within this feasible region is 110, which is the correct answer.

The maximum profit can be calculated by finding the maximum quantity of mangoes and bananas that can be bought within the available capital and weight limit of the cart.

Let's assume the quantity of mangoes bought is x kg and the quantity of bananas bought is y kg.

The cost of buying x kg of mangoes is 8,000x, and the cost of buying y kg of bananas is 6,000y.

We know that the total cost of buying mangoes and bananas cannot exceed the available capital, so we have the equation: 8,000x + 6,000y ≤ 1,200,000.

We also know that the total weight of mangoes and bananas cannot exceed 180 kg, so we have the equation: x + y ≤ 180.

To maximize the profit, we need to maximize the selling price of mangoes and bananas.

The selling price of x kg of mangoes is 9,200x, and the selling price of y kg of bananas is 7,000y.

The profit can be calculated as: Profit = (9,200x + 7,000y) - (8,000x + 6,000y)

To find the maximum profit, we need to solve the system of inequalities:
8,000x + 6,000y ≤ 1,200,000
x + y ≤ 180

By solving this system, we find that the maximum profit is Rp 192,000.

The given mathematical model has two inequality constraints: 3x + 2y

The correct answer is D because it correctly represents the constraints of the problem. The inequality 2x + y ≤ 100 represents the constraint on the number of tables that can be produced, as each table requires 10 pieces of wood (2x) and 1 chair requires 5 pieces of wood (y). The inequality 5x + 2y ≤ 50 represents the constraint on the number of chairs that can be produced. The constraints x ≥ 0 and y ≥ 0 ensure that the number of tables and chairs produced cannot be negative.

The maximum profit that can be obtained from selling the houses can be calculated by finding the maximum number of houses that can be built on the given land area and then multiplying it by the profit per unit for each type of house.

Since 100m2 is required for type A house and 75m2 is required for type B house, the maximum number of type A houses that can be built is 10,000m2 / 100m2 = 100 units. Similarly, the maximum number of type B houses that can be built is 10,000m2 / 75m2 = 133.33 units.

However, the maximum number of houses that can be built is limited to 125 units. Therefore, the maximum profit can be obtained by building 100 units of type A houses and 25 units of type B houses.

The profit from selling type A houses is Rp 6,000,000.00 x 100 units = Rp 600,000,000.00.
The profit from selling type B houses is Rp 4,000,000.00 x 25 units = Rp 100,000,000.00.

Therefore, the maximum profit that can be obtained from selling the houses is Rp 600,000,000.00 + Rp 100,000,000.00 = Rp 700,000,000.00.

Hence, the correct answer is a.Rp 550.000.000,00.