Graph the system of constraints and find the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that maximize the objective function.

Constraints:
[tex]\[
\begin{cases}
x \geq 0 \\
y \geq 0 \\
y \leq \frac{1}{5}x + 2 \\
y + x \leq 5
\end{cases}
\][/tex]

Objective function: [tex]\(C = 7x - 3y\)[/tex]

Possible solutions:
A. [tex]\((2.5, 2.5)\)[/tex]
B. [tex]\((0, 2)\)[/tex]
C. [tex]\((0, 0)\)[/tex]
D. [tex]\((5, 0)\)[/tex]



Answer :

To solve this problem, let’s follow these steps:

1. Graph the constraints: We will graph the feasible region determined by the constraints.
2. Identify the corner points: Determine the intersection points of the constraints.
3. Evaluate the objective function at the corner points: Calculate the value of the objective function [tex]\(C = 7x - 3y\)[/tex] at each corner point.
4. Determine the maximum value: Identify which point gives the maximum value for the objective function [tex]\(C\)[/tex].

### Step 1: Graph the Constraints

The given constraints are:
1. [tex]\(x \geq 0\)[/tex]
2. [tex]\(y \geq 0\)[/tex]
3. [tex]\(y \leq \frac{1}{5}x + 2\)[/tex]
4. [tex]\(y + x \leq 5\)[/tex]

### Step 2: Identify the Intersection Points

To find the intersection points, we solve the equations derived from the constraints:

#### Intersection of [tex]\( y = \frac{1}{5}x + 2 \)[/tex] and [tex]\( y + x = 5 \)[/tex]:

Set [tex]\(y = \frac{1}{5}x + 2\)[/tex] into [tex]\( y + x = 5\)[/tex]:

[tex]\[ \frac{1}{5}x + 2 + x = 5 \][/tex]
[tex]\[ x + \frac{1}{5}x = 3 \][/tex]
[tex]\[ \frac{6}{5}x = 3 \][/tex]
[tex]\[ x = \frac{5 \cdot 3}{6} = 2.5 \][/tex]
[tex]\[ y = \frac{1}{5}(2.5) + 2 = 2.5 \][/tex]

So, one of the intersection points is [tex]\((2.5, 2.5)\)[/tex].

#### Other points from constraints:

- [tex]\( (0, 2) \)[/tex] : Intersection of [tex]\( y = \frac{1}{5}x + 2 \)[/tex] and [tex]\(x = 0 \)[/tex]
- [tex]\( (5, 0) \)[/tex] : Intersection of [tex]\( y + x = 5 \)[/tex] and [tex]\(y = 0 \)[/tex]
- [tex]\( (0, 0) \)[/tex] : Intersection of [tex]\( y = 0 \)[/tex] and [tex]\(x = 0 \)[/tex]

So, the corner points of the feasible region are:
[tex]\[ (2.5, 2.5), (0, 2), (0, 0), (5, 0) \][/tex]

### Step 3: Evaluate the Objective Function at Each Corner Point

We now calculate the value of the objective function [tex]\(C = 7x - 3y\)[/tex] at each corner point:

- At [tex]\((2.5, 2.5)\)[/tex]:
[tex]\[ C = 7(2.5) - 3(2.5) = 17.5 - 7.5 = 10 \][/tex]

- At [tex]\((0, 2)\)[/tex]:
[tex]\[ C = 7(0) - 3(2) = 0 - 6 = -6 \][/tex]

- At [tex]\((0, 0)\)[/tex]:
[tex]\[ C = 7(0) - 3(0) = 0 \][/tex]

- At [tex]\((5, 0)\)[/tex]:
[tex]\[ C = 7(5) - 3(0) = 35 - 0 = 35 \][/tex]

### Step 4: Determine the Maximum Value

Comparing the values obtained:
[tex]\[ 10, -6, 0, 35 \][/tex]

The maximum value of the objective function is 35 which occurs at the point [tex]\((5, 0)\)[/tex].

### Conclusion

Therefore, the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that maximize the objective function [tex]\(C = 7x - 3y\)[/tex] within the given constraints are:
[tex]\[ x = 5 \quad \text{and} \quad y = 0 \][/tex]

Thus, the maximum value of the objective function is 35 at the point [tex]\((5, 0)\)[/tex].