Z = 20x1 + 202, subject to x1 ≥ 0, x2 ≥ 0, x1 + 2x2 ≥ 8, 3x1 + 2x2 ≥ 15, 5x1 + 2x2 ≥ 20. The minimum value of Z occurs at

The correct option is (7/2, 9/4).

To find the point at which the objective function $Z=20x_1+20x_2$ is minimized, we need to analyze the system of linear inequalities (constraints) and determine the corner points of the feasible region.

The given constraints are:
1. $x_1\ge 0,x_2\ge 0$ (First quadrant)
2. $x_1+2x_2\ge 8$
3. $3x_1+2x_2\ge 15$
4. $5x_1+2x_2\ge 20$

Let's find the intersection points of the boundary lines to locate the corner points of the feasible region:

Boundary Line 1: $x_1+2x_2=8$
If $x_1=0$, then $x_2=4$. Point: (0, 4)
If $x_2=0$, then $x_1=8$. Point: (8, 0)

Boundary Line 2: $3x_1+2x_2=15$
If $x_1=0$, then $x_2=15/2=7.5$. Point: (0, 15/2)
If $x_2=0$, then $x_1=5$. Point: (5, 0)

Boundary Line 3: $5x_1+2x_2=20$
If $x_1=0$, then $x_2=10$. Point: (0, 10)
If $x_2=0$, then $x_1=4$. Point: (4, 0)

Since all constraints are of the type "$\ge$", the feasible region is unbounded and lies above and to the right of the boundary lines in the first quadrant.

Let's determine the intersection points between these boundary lines to identify the corner points of this boundary:

Intersection of Line 1 ($x_1+2x_2=8$) and Line 2 ($3x_1+2x_2=15$):
Subtracting the first equation from the second:
$(3x_1+2x_2)-(x_1+2x_2)=15-8$
$2x_1=7\Rightarrow x_1=7/2$
Substituting $x_1=7/2$ into Line 1:
$7/2+2x_2=8\Rightarrow 2x_2=8-7/2=9/2\Rightarrow x_2=9/4$
This gives the corner point (7/2, 9/4).

Intersection of Line 2 ($3x_1+2x_2=15$) and Line 3 ($5x_1+2x_2=20$):
Subtracting the first equation from the second:
$(5x_1+2x_2)-(3x_1+2x_2)=20-15$
$2x_1=5\Rightarrow x_1=5/2$
Substituting $x_1=5/2$ into Line 2:
$3(5/2)+2x_2=15\Rightarrow 15/2+2x_2=15\Rightarrow 2x_2=15/2\Rightarrow x_2=15/4$
This gives the corner point (5/2, 15/4).

The corner points of the feasible region are:
- $(8,0)$
- $(7/2,9/4)$
- $(5/2,15/4)$
- $(0,10)$

Now, we evaluate the objective function $Z=20x_1+20x_2$ at each corner point:

1. At $(8,0)$:
$Z=20\left(8\right)+20\left(0\right)=160$

2. At $(7/2,9/4)$:
$Z=20(7/2)+20(9/4)=70+45=115$

3. At $(5/2,15/4)$:
$Z=20(5/2)+20(15/4)=50+75=125$

4. At $(0,10)$:
$Z=20\left(0\right)+20\left(10\right)=200$

Comparing the values, the minimum value of $Z$ is $115$, which occurs at the corner point $(7/2,9/4)$.