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Zorluk: OrtaSystems of Linear Inequalities in Two Variables

A point (x,y)(x, y) in the coordinate plane satisfies the system of inequalities below.

y2x4y \geq 2x - 4
yx+5y \leq -x + 5
x0x \geq 0
y0y \geq 0

What is the maximum possible value of xx for this point?

Cevap: 3

Cevap

The maximum possible value of xx that satisfies the system of inequalities is 3.
To find the maximum possible value of xx for a point satisfying the system of inequalities, we identify the vertices of the region. The upper-right boundary of the region is formed by the intersection of the lines y=2x4y = 2x - 4 and y=x+5y = -x + 5. Solving 2x4=x+52x - 4 = -x + 5 gives 3x=9    x=33x = 9 \implies x = 3. At this point, y=3+5=2y = -3 + 5 = 2, which satisfies the constraints x0x \geq 0 and y0y \geq 0. The other boundary vertices of the region are (0,0)(0, 0), (2,0)(2, 0), and (0,5)(0, 5). Comparing the x-coordinates of these vertices (00, 22, and 33), we see that the maximum possible value of xx is 33.

Adım Adım Çözüm

1
Find the intersection point of the boundary lines y=2x4y = 2x - 4 and y=x+5y = -x + 5.
x=3x = 3
Setting the two boundary line equations equal to each other (2x4=x+52x - 4 = -x + 5) allows us to find the x-coordinate where the boundaries cross.
2
Substitute x=3x = 3 back into one of the boundary equations to find the y-coordinate.
y=2y = 2
This yields the intersection point (3,2)(3, 2) which lies on both boundary lines.
3
Verify that the intersection point (3,2)(3, 2) satisfies the other constraints: x0x \geq 0 and y0y \geq 0.
303 \geq 0 and 202 \geq 0 (both true)
The point must lie within the first quadrant to be a valid solution.
4
Determine the remaining boundary vertices of the solution region in the first quadrant.
Vertices are (0,0)(0, 0), (0,5)(0, 5), (2,0)(2, 0), and (3,2)(3, 2).
Comparing all vertices will confirm if (3,2)(3, 2) indeed provides the maximum value of xx.

Anahtar Kavram

Systems of Linear Inequalities in Two Variables
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