Answer :
To find the maximum profit for the profit function [tex]\( P = 2x + 5y \)[/tex] subject to the constraints:
[tex]\[ \begin{cases} 2x + y \leq 80 \\ x + 2y \leq 80 \\ x \geq 0 \\ y \geq 0 \end{cases} \][/tex]
we'll follow these steps:
### Step 1: Understand the constraints
We need to visualize the constraints on a graph and find their intersections to determine the feasible region.
- The first constraint [tex]\(2x + y \leq 80\)[/tex]:
- If [tex]\(x = 0\)[/tex], then [tex]\(y \leq 80\)[/tex].
- If [tex]\(y = 0\)[/tex], then [tex]\(x \leq 40\)[/tex].
- Intersection points can be found by setting one variable to 0 and solving for the other.
- The second constraint [tex]\(x + 2y \leq 80\)[/tex]:
- If [tex]\(x = 0\)[/tex], then [tex]\(2y \leq 80 \Rightarrow y \leq 40\)[/tex].
- If [tex]\(y = 0\)[/tex], then [tex]\(x \leq 80\)[/tex].
- The non-negativity constraints [tex]\(x \geq 0\)[/tex] and [tex]\(y \geq 0\)[/tex] denote that the solutions must be in the first quadrant.
### Step 2: Identify intersection points
To find the vertices of the feasible region, we solve the system of linear equations given by the constraints:
1. Intersection of [tex]\(2x + y = 80\)[/tex] and [tex]\(x + 2y = 80\)[/tex]:
Solving this system of equations:
[tex]\[ \begin{cases} 2x + y = 80 \\ x + 2y = 80 \end{cases} \][/tex]
Multiply the second equation by 2 to eliminate one of the variables:
[tex]\[ \begin{cases} 2x + y = 80 \\ 2x + 4y = 160 \end{cases} \][/tex]
Subtract the first equation from the second:
[tex]\[ 2x + 4y - (2x + y) = 160 - 80 \\ 3y = 80 \\ y = \frac{80}{3} \][/tex]
Substitute [tex]\( y = \frac{80}{3} \)[/tex] back into the first equation [tex]\( 2x + \frac{80}{3} = 80 \)[/tex]:
[tex]\[ 2x = 80 - \frac{80}{3} \\ 2x = \frac{240}{3} - \frac{80}{3} \\ 2x = \frac{160}{3} \\ x = \frac{80}{3} \][/tex]
So, one vertex is [tex]\( \left(\frac{80}{3}, \frac{80}{3}\right) \)[/tex].
2. Intersection of [tex]\(2x + y = 80\)[/tex] with the x-axis (where [tex]\( y = 0 \)[/tex]):
Set [tex]\( y = 0 \)[/tex] in the first constraint:
[tex]\[ 2x = 80 \\ x = 40 \][/tex]
So, another vertex is [tex]\( (40, 0) \)[/tex].
3. Intersection of [tex]\(x + 2y = 80\)[/tex] with the y-axis (where [tex]\( x = 0 \)[/tex]):
Set [tex]\( x = 0 \)[/tex] in the second constraint:
[tex]\[ 2y = 80 \\ y = 40 \][/tex]
So, another vertex is [tex]\( (0, 40) \)[/tex].
### Step 3: Calculate the profit at each vertex
Evaluate the profit function [tex]\( P = 2x + 5y \)[/tex] at each vertex of the feasible region:
1. At [tex]\( \left(\frac{80}{3}, \frac{80}{3}\right) \)[/tex]:
[tex]\[ P = 2 \cdot \frac{80}{3} + 5 \cdot \frac{80}{3} = \frac{160}{3} + \frac{400}{3} = \frac{560}{3} \approx 186.67 \][/tex]
2. At [tex]\( (40, 0) \)[/tex]:
[tex]\[ P = 2 \cdot 40 + 5 \cdot 0 = 80 \][/tex]
3. At [tex]\( (0, 40) \)[/tex]:
[tex]\[ P = 2 \cdot 0 + 5 \cdot 40 = 200 \][/tex]
### Step 4: Determine the maximum profit
By evaluating the profit function at all vertices, we observe that the maximum profit occurs at [tex]\( (0, 40) \)[/tex] and the maximum profit is 200.
Therefore, the optimal solution is [tex]\( x = 0 \)[/tex] and [tex]\( y = 40 \)[/tex], yielding a maximum profit of 200.
[tex]\[ \begin{cases} 2x + y \leq 80 \\ x + 2y \leq 80 \\ x \geq 0 \\ y \geq 0 \end{cases} \][/tex]
we'll follow these steps:
### Step 1: Understand the constraints
We need to visualize the constraints on a graph and find their intersections to determine the feasible region.
- The first constraint [tex]\(2x + y \leq 80\)[/tex]:
- If [tex]\(x = 0\)[/tex], then [tex]\(y \leq 80\)[/tex].
- If [tex]\(y = 0\)[/tex], then [tex]\(x \leq 40\)[/tex].
- Intersection points can be found by setting one variable to 0 and solving for the other.
- The second constraint [tex]\(x + 2y \leq 80\)[/tex]:
- If [tex]\(x = 0\)[/tex], then [tex]\(2y \leq 80 \Rightarrow y \leq 40\)[/tex].
- If [tex]\(y = 0\)[/tex], then [tex]\(x \leq 80\)[/tex].
- The non-negativity constraints [tex]\(x \geq 0\)[/tex] and [tex]\(y \geq 0\)[/tex] denote that the solutions must be in the first quadrant.
### Step 2: Identify intersection points
To find the vertices of the feasible region, we solve the system of linear equations given by the constraints:
1. Intersection of [tex]\(2x + y = 80\)[/tex] and [tex]\(x + 2y = 80\)[/tex]:
Solving this system of equations:
[tex]\[ \begin{cases} 2x + y = 80 \\ x + 2y = 80 \end{cases} \][/tex]
Multiply the second equation by 2 to eliminate one of the variables:
[tex]\[ \begin{cases} 2x + y = 80 \\ 2x + 4y = 160 \end{cases} \][/tex]
Subtract the first equation from the second:
[tex]\[ 2x + 4y - (2x + y) = 160 - 80 \\ 3y = 80 \\ y = \frac{80}{3} \][/tex]
Substitute [tex]\( y = \frac{80}{3} \)[/tex] back into the first equation [tex]\( 2x + \frac{80}{3} = 80 \)[/tex]:
[tex]\[ 2x = 80 - \frac{80}{3} \\ 2x = \frac{240}{3} - \frac{80}{3} \\ 2x = \frac{160}{3} \\ x = \frac{80}{3} \][/tex]
So, one vertex is [tex]\( \left(\frac{80}{3}, \frac{80}{3}\right) \)[/tex].
2. Intersection of [tex]\(2x + y = 80\)[/tex] with the x-axis (where [tex]\( y = 0 \)[/tex]):
Set [tex]\( y = 0 \)[/tex] in the first constraint:
[tex]\[ 2x = 80 \\ x = 40 \][/tex]
So, another vertex is [tex]\( (40, 0) \)[/tex].
3. Intersection of [tex]\(x + 2y = 80\)[/tex] with the y-axis (where [tex]\( x = 0 \)[/tex]):
Set [tex]\( x = 0 \)[/tex] in the second constraint:
[tex]\[ 2y = 80 \\ y = 40 \][/tex]
So, another vertex is [tex]\( (0, 40) \)[/tex].
### Step 3: Calculate the profit at each vertex
Evaluate the profit function [tex]\( P = 2x + 5y \)[/tex] at each vertex of the feasible region:
1. At [tex]\( \left(\frac{80}{3}, \frac{80}{3}\right) \)[/tex]:
[tex]\[ P = 2 \cdot \frac{80}{3} + 5 \cdot \frac{80}{3} = \frac{160}{3} + \frac{400}{3} = \frac{560}{3} \approx 186.67 \][/tex]
2. At [tex]\( (40, 0) \)[/tex]:
[tex]\[ P = 2 \cdot 40 + 5 \cdot 0 = 80 \][/tex]
3. At [tex]\( (0, 40) \)[/tex]:
[tex]\[ P = 2 \cdot 0 + 5 \cdot 40 = 200 \][/tex]
### Step 4: Determine the maximum profit
By evaluating the profit function at all vertices, we observe that the maximum profit occurs at [tex]\( (0, 40) \)[/tex] and the maximum profit is 200.
Therefore, the optimal solution is [tex]\( x = 0 \)[/tex] and [tex]\( y = 40 \)[/tex], yielding a maximum profit of 200.
