Minimize

[tex]\[
\begin{array}{l}
z = 2x + y \\
2y + 3x \geq 16 \\
5y + 4x \geq 32
\end{array}
\][/tex]

Subject to

[tex]\[
\begin{array}{ll}
y + x & \geq 7 \\
x & \geq 0 \\
y & \geq 0
\end{array}
\][/tex]

Minimum is [tex]$\square$[/tex]



Answer :

To minimize the objective function [tex]\( z = 2x + y \)[/tex] subject to the given constraints:

1. [tex]\( 2y + 3x \geq 16 \)[/tex]
2. [tex]\( 5y + 4x \geq 32 \)[/tex]
3. [tex]\( y + x \geq 7 \)[/tex]
4. [tex]\( x \geq 0 \)[/tex]
5. [tex]\( y \geq 0 \)[/tex]

we follow a systematic approach.

### Step 1: Define the Constraints

First, let's rewrite the inequalities in standard form to understand the feasible region:

- [tex]\( 2y + 3x \geq 16 \)[/tex]
- [tex]\( 5y + 4x \geq 32 \)[/tex]
- [tex]\( y + x \geq 7 \)[/tex]

Additionally, we have non-negativity constraints: [tex]\( x \geq 0 \)[/tex] and [tex]\( y \geq 0 \)[/tex].

### Step 2: Find the Feasible Region

The feasible region is defined by the intersection of the half-planes formed by these inequalities.

### Step 3: Determine Corner Points of the Feasible Region

To minimize the objective function linearly, we need to find the points where the constraint lines intersect. Let's identify some of these points:

1. Intersection of [tex]\( 2y + 3x = 16 \)[/tex] and [tex]\( 5y + 4x = 32 \)[/tex]
2. Intersection of [tex]\( 2y + 3x = 16 \)[/tex] and [tex]\( y + x = 7 \)[/tex]
3. Intersection of [tex]\( 5y + 4x = 32 \)[/tex] and [tex]\( y + x = 7 \)[/tex]
4. Check intersections with the axes, since non-negativity bounds might also be relevant.

### Step 4: Compute the Objective Function

Having identified potential intersection points, we calculate the value of [tex]\( z = 2x + y \)[/tex] for each point and compare them to find the minimum value:

1. Evaluate [tex]\( z = 2x + y \)[/tex] at the feasible points.
2. Identify the point which yields the minimum value of [tex]\( z \)[/tex].

### Step 5: Verify and Confirm

After evaluation, the point that minimizes the objective function [tex]\( z = 2x + y \)[/tex] within the feasible region is found to be [tex]\( (x, y) = (0, 8) \)[/tex], and the minimum value of [tex]\( z \)[/tex] at this point is:

[tex]\[ z_{\text{min}} = 2(0) + 8 = 8 \][/tex]

Thus, the minimum value of [tex]\( z \)[/tex] is [tex]\(\boxed{8}\)[/tex].