`Z=20x_1+20x_2`, subject to `x_1 le 0, x_2 le 0, x_1+2x_2 ge 8, 3x_1 +2x_2 ge 15, 5x_1+2x_2 ge 20`.The minimum value of Z occurs at :

Options

A. (8,0)
B. `(5/2 , 15/4)`
C. `(7/2 , 9/4)`
D. (0 , 10)

Answer

To solve the problem of minimizing \( Z = 20x_1 + 20x_2 \) subject to the constraints:

1. \( x_1 \leq 0 \)
2. \( x_2 \leq 0 \)
3. \( x_1 + 2x_2 \geq 8 \)
4. \( 3x_1 + 2x_2 \geq 15 \)
5. \( 5x_1 + 2x_2 \geq 20 \)

we will follow these steps:

### Step 1: Identify the Constraints
We have the following constraints:
- \( x_1 \leq 0 \)
- \( x_2 \leq 0 \)
- \( x_1 + 2x_2 \geq 8 \)
- \( 3x_1 + 2x_2 \geq 15 \)
- \( 5x_1 + 2x_2 \geq 20 \)

### Step 2: Convert Inequalities to Equations
To find the intersection points, we convert the inequalities into equalities:
1. \( x_1 + 2x_2 = 8 \) (Equation 1)
2. \( 3x_1 + 2x_2 = 15 \) (Equation 2)
3. \( 5x_1 + 2x_2 = 20 \) (Equation 3)

### Step 3: Find Intersection Points
We will solve pairs of equations to find the points of intersection.

#### Intersection of Equation 1 and Equation 2:
\[ x_1 + 2x_2 = 8 \quad (1) \]
\[ 3x_1 + 2x_2 = 15 \quad (2) \]

Subtract (1) from (2):
\[ (3x_1 + 2x_2) - (x_1 + 2x_2) = 15 - 8 \]
\[ 2x_1 = 7 \implies x_1 = \frac{7}{2} \]

Substituting \( x_1 \) back into Equation (1):
\[ \frac{7}{2} + 2x_2 = 8 \implies 2x_2 = 8 - \frac{7}{2} = \frac{16}{2} - \frac{7}{2} = \frac{9}{2} \implies x_2 = \frac{9}{4} \]

So, one intersection point is \( \left(\frac{7}{2}, \frac{9}{4}\right) \).

#### Intersection of Equation 2 and Equation 3:
\[ 3x_1 + 2x_2 = 15 \quad (2) \]
\[ 5x_1 + 2x_2 = 20 \quad (3) \]

Subtract (2) from (3):
\[ (5x_1 + 2x_2) - (3x_1 + 2x_2) = 20 - 15 \]
\[ 2x_1 = 5 \implies x_1 = \frac{5}{2} \]

Substituting \( x_1 \) back into Equation (2):
\[ 3\left(\frac{5}{2}\right) + 2x_2 = 15 \implies \frac{15}{2} + 2x_2 = 15 \]
\[ 2x_2 = 15 - \frac{15}{2} = \frac{30}{2} - \frac{15}{2} = \frac{15}{2} \implies x_2 = \frac{15}{4} \]

So, another intersection point is \( \left(\frac{5}{2}, \frac{15}{4}\right) \).

#### Intersection of Equation 1 and Equation 3:
\[ x_1 + 2x_2 = 8 \quad (1) \]
\[ 5x_1 + 2x_2 = 20 \quad (3) \]

Subtract (1) from (3):
\[ (5x_1 + 2x_2) - (x_1 + 2x_2) = 20 - 8 \]
\[ 4x_1 = 12 \implies x_1 = 3 \]

Substituting \( x_1 \) back into Equation (1):
\[ 3 + 2x_2 = 8 \implies 2x_2 = 5 \implies x_2 = \frac{5}{2} \]

So, the third intersection point is \( (3, \frac{5}{2}) \).

### Step 4: Evaluate the Objective Function at Each Intersection Point
Now we will evaluate \( Z = 20x_1 + 20x_2 \) at each intersection point.

1. For \( \left(\frac{7}{2}, \frac{9}{4}\right) \):
\[ Z = 20\left(\frac{7}{2}\right) + 20\left(\frac{9}{4}\right) = 70 + 45 = 115 \]

2. For \( \left(\frac{5}{2}, \frac{15}{4}\right) \):
\[ Z = 20\left(\frac{5}{2}\right) + 20\left(\frac{15}{4}\right) = 50 + 75 = 125 \]

3. For \( (3, \frac{5}{2}) \):
\[ Z = 20(3) + 20\left(\frac{5}{2}\right) = 60 + 50 = 110 \]

### Step 5: Determine the Minimum Value
The minimum value of \( Z \) occurs at the point \( (3, \frac{5}{2}) \) with:
\[ \text{Minimum } Z = 110 \]

### Final Answer
The minimum value of \( Z \) occurs at \( (3, \frac{5}{2}) \).