Program Linier 01

2.

1. Nilai optimum z = 5x + 2y dari model matematika berikut: 3x + 2y < 36.000 x + 2y < 20.000 x > 0 y > 0adalah

A. 20.000
B. 52.000
C. 60.000
D. 86.000
E. 100.000

Correct Answer

A. B. 52.000

Explanation

The given mathematical model has two inequality constraints: 3x + 2y < 36,000 and x + 2y < 20,000. The objective function is z = 5x + 2y. To find the optimum value of z, we need to find the values of x and y that satisfy the constraints and maximize the objective function. By solving the system of inequalities, we can find that x = 8,000 and y = 6,000. Plugging these values into the objective function, we get z = 5(8,000) + 2(6,000) = 52,000. Therefore, the correct answer is b. 52,000.

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3.

. Nilai optimum z = 5x + 3y pada model matematika berikut: 2x + 3y < 80 5x + y < 70 x > 0, y > 0 adalah daerah yang ditunjukkan oleh

A.90
B. 105
C.100
D.110
E.130

Correct Answer

A. D.110

Explanation

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.

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4.

Luas daerah parkir 1.760 m2. Luas rata – rata untuk mobil kecil 4 m2 dan mobil besar 20 m2.Daya tampung maksimum hanya 200 kendaraan, biaya parkir mobil kecil Rp 1.000,00/jam danmobil besar Rp 2.000,00/jam. Jika dalam satu jam terisi penuh dan tidak kendaraan yangpergi dan datang, maka hasil maksimum tempat parkir itu adalah

a.Rp 176.000,00
B.Rp 200.000,00
C.Rp 260.000,00
D.Rp 300.000,00
E.Rp 340.000,00

Correct Answer

A. C.Rp 260.000,00

Explanation

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.

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5.

Seorang pedagang menjual buah mangga dan pisang dengan menggunakan gerobak.Pedagang tersebut membeli mangga dengan harga Rp. 8.000,00/kg dan pisang Rp 6.000,00/kg. Modal yang tersedia Rp 1.200.000,00 dan gerobaknya hanya dapat memuat mangga danpisang sebanyak 180 kg. Jika harga jual mangga Rp 9.200,00/kg dan pisang Rp 7.000,00/kg,maka laba maksimum yang diperoleh adalah …

a.Rp 150.000,00
B.Rp 180.000,00
C.Rp 192.000,00
D.Rp 204.000,00
E.Rp 216.000,00

Correct Answer

A. C.Rp 192.000,00

Explanation

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.

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1

6.

Tanah seluas 10.000 m2 akan dibangun rumah tipe A dan tipe B. Untuk tipe A diperlukan 100m2 dan dan tipe B diperlukan 75 m2. Jumlah rumah yang akan dibangun paling banyak 125unit. Keuntungan rumah tipe A adalah Rp 6.000.000,00/unit dan tipe B adalah Rp4.000.000,00/unit. Keuntungan maksimum yang dapat diperoleh daru penjualan rumahtersebut adalah

A.Rp 550.000.000,00
B.Rp 600.000.000,00
C.Rp 700.000.000,00
D.Rp 800.000.000,00
E.Rp 900.000.000,00

Correct Answer

A. A.Rp 550.000.000,00

Explanation

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.

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7.

Seorang pengusaha mebel akan memproduksi meja dan kursi yang menggunakan bahan dari papan-papan kayu dengan ukuran tertentu. Satu meja memerlukan bahan 10 potong dan satu kursi memerlukan 5 potong papan. Papan yang tersedia ada 500 potong. Biaya pembuatan satu meja Rp100.000,00 dan biaya pembuatan satu kursi Rp40.000,00. Anggaran yang tersedia Rp1.000.000,00. Model matematika dari persoalan tersebut adalah

A. .x + 2y ≤ 100; 5x + 2y ≤ 50; x ≥ 0; y ≥ 0
B. x + 2y ≤ 100; 2x + 5y ≤ 50; x ≥ 0; y ≥ 0
C. 2x + y ≤ 100; 2x + 5y ≤ 50; x ≥ 0; y ≥ 0
D. 2x + y ≤ 100; 5x + 2y ≤ 50; x ≥ 0; y ≥ 0 D.2x + y ≤ 100; 5x + 2y ≤ 50; x ≥ 0; y ≥ 0 D.2x + y ≤ 100; 5x + 2y ≤ 50; x ≥ 0; y ≥ 0
E. 2x + y ≥ 100; 5x + 2y ≥ 50; x ≥ 0; y ≥ 0

Correct Answer

A. D. 2x + y ≤ 100; 5x + 2y ≤ 50; x ≥ 0; y ≥ 0 D.2x + y ≤ 100; 5x + 2y ≤ 50; x ≥ 0; y ≥ 0 D.2x + y ≤ 100; 5x + 2y ≤ 50; x ≥ 0; y ≥ 0

Explanation

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.

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Program Linier 01

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1. Nilai optimum z = 5x + 2y dari model matematika berikut: 3x + 2y < 36.000 x + 2y < 20.000 x > 0 y > 0adalah

. Nilai optimum z = 5x + 3y pada model matematika berikut: 2x + 3y < 80 5x + y < 70 x > 0, y > 0 adalah daerah yang ditunjukkan oleh