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Tentukan daeran arsiran dari Sistem RertidakSamaan linier berikut ret \( 2 y \leq 10 \) \( x+y \leq 7 \) \( x \geq 0 \) \( y \geq 0 \)

Ask by Campbell Campos. in Indonesia
Feb 25,2025

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The shaded region includes all points (x, y) where: - x is greater than or equal to 0, - y is greater than or equal to 0, - y is less than or equal to 5, - and the sum of x and y is less than or equal to 7. This forms a quadrilateral with vertices at (0,0), (7,0), (2,5), and (0,5).

Solution

We are given the following system of inequalities:   1. 2y ≤ 10   2. x + y ≤ 7   3. x ≥ 0   4. y ≥ 0 Step 1. Rewrite the inequalities in a simpler form. • From 2y ≤ 10, divide both sides by 2:   y ≤ 5 Thus, the system becomes:   y ≤ 5   x + y ≤ 7   x ≥ 0   y ≥ 0 Step 2. Interpret the constraints. • y ≤ 5 means that all points must have a y‑coordinate less than or equal to 5 (horizontal line at y = 5). • x + y ≤ 7 indicates the region below (or on) the line x + y = 7. • x ≥ 0 and y ≥ 0 confine the region to the first quadrant. Step 3. Identify the key boundary lines and their intersections. The boundaries of the feasible region (the “shaded region”) are defined by:   a) x = 0 (y‑axis),   b) y = 0 (x‑axis),   c) y = 5, and   d) x + y = 7. Find the intersection points (vertices) of the region: 1. Intersection of x + y = 7 and y = 0:   Substitute y = 0 into x + y = 7 → x = 7.   Vertex: (7, 0). 2. Intersection of x + y = 7 and y = 5:   Substitute y = 5: x + 5 = 7 → x = 2.   Vertex: (2, 5). 3. Intersection of y = 5 and x = 0:   Vertex: (0, 5). 4. Intersection of x = 0 and y = 0:   Vertex: (0, 0). Step 4. Describe the feasible (shaded) region. The feasible region is the set of all points (x,y) in the first quadrant that satisfy:   x ≥ 0, y ≥ 0, y ≤ 5, and x + y ≤ 7. The region is bounded by the vertices (0, 0), (7, 0), (2, 5), and (0, 5). It is a quadrilateral where:  • The segment from (0, 0) to (7, 0) lies along the x‑axis,  • The segment from (7, 0) to (2, 5) lies along the line x + y = 7,  • The segment from (2, 5) to (0, 5) lies along the horizontal line y = 5, and  • The segment from (0, 5) back to (0, 0) lies along the y‑axis. Thus, the shaded region is the quadrilateral with vertices (0, 0), (7, 0), (2, 5), and (0, 5). Final Answer: The feasible (shaded) region for the system is:  {x, y ∈ ℝ : x ≥ 0, y ≥ 0, y ≤ 5, and x + y ≤ 7}, which is the quadrilateral with vertices (0, 0), (7, 0), (2, 5), and (0, 5).

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Beyond the Answer

Sistem pertidaksamaan linear yang diberikan dapat divisualisasikan dalam bidang Kartesius. Pertama, untuk \(2y \leq 10\), kita bisa menyederhanakannya menjadi \(y \leq 5\), yang menunjukkan garis horizontal pada \(y = 5\). Kedua, dari \(x + y \leq 7\), dapat digambar garis \(y = 7 - x\), yang memiliki intercept pada \(y\) di 7 dan intercept pada \(x\) di 7 juga. Daerah arsiran yang memenuhi semua pertidaksamaan tersebut akan berupa segitiga yang terletak di kuadran pertama, di bawah garis \(y = 5\) dan di bawah garis \(y = 7 - x\). Pastikan untuk mengarsir daerah yang termasuk di dalam batasan tersebut, yaitu titik-titik yang memenuhi keempat pertidaksamaan untuk mendapatkan solusi yang sah.

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