In a linear programming problem, the constraints on decision variables x and y are y-2x ≤ 0, y ≥ 0, 0 ≤ x ≤ 5. The feasible region of the above…

We need to find the feasible region defined by these constraints:

$$y - 2x \leq 0 \implies y \leq 2x$$y−2x≤0⟹y≤2x

$$y \geq 0$$y≥0

$$0 \leq x \leq 5$$0≤x≤5

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$y \leq 2x$y≤2x means the region below (or on) the line $y = 2x$y=2x, which passes through the origin with slope $2$2.

$y \geq 0$y≥0 means we stay on or above the $x$x-axis.

$0 \leq x \leq 5$0≤x≤5 means we stay between the $y$y-axis and the vertical line $x = 5$x=5.

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The region where all constraints overlap is bounded by three corner points, found by intersecting the boundary lines:

At $x = 0$x=0 and $y = 0$y=0:

$$A = (0,\, 0)$$A=(0,0)

At $x = 5$x=5 and $y = 0$y=0:

$$B = (5,\, 0)$$B=(5,0)

At $x = 5$x=5 and $y = 2x$y=2x:

$$y = 2(5) = 10$$y=2(5)=10

$$C = (5,\, 10)$$C=(5,10)

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Connecting $(0,\, 0)$(0,0), $(5,\, 0)$(5,0), and $(5,\, 10)$(5,10) gives us a triangle with:

Base along the $x$x-axis $= 5$=5

Vertical side from $(5,\, 0)$(5,0) to $(5,\, 10)$(5,10) $= 10$=10

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The feasible region is a triangle with vertices at $(0,\, 0)$(0,0), $(5,\, 0)$(5,0), and $(5,\, 10)$(5,10).