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Sagot :
To find the maximum value of [tex]\(P = 4x + 2y\)[/tex] given the constraints:
[tex]\[ \begin{array}{l} x + 2y \leq 10 \\ y \leq 2 \\ x \geq 0 \\ y \geq 0 \end{array} \][/tex]
we need to solve this as a linear programming problem. Let’s follow these steps:
1. Identify the objective function and constraints:
- Objective function: [tex]\(P = 4x + 2y\)[/tex]
- Constraints:
[tex]\[ \begin{array}{l} x + 2y \leq 10 \\ y \leq 2 \\ x \geq 0 \\ y \geq 0 \end{array} \][/tex]
2. Graph the constraints:
- Plot the line [tex]\(x + 2y = 10\)[/tex]. This will be a boundary line.
- Plot the horizontal line [tex]\(y = 2\)[/tex].
- Add the non-negativity constraints [tex]\(x \geq 0\)[/tex] and [tex]\(y \geq 0\)[/tex].
3. Determine the feasible region:
- The feasible region is the area where all constraints overlap and it lies in the first quadrant (both [tex]\(x\)[/tex] and [tex]\(y\)[/tex] are non-negative).
4. Identify the corner points of the feasible region:
- The boundary line [tex]\(x + 2y = 10\)[/tex] intersects the [tex]\(y\)[/tex]-axis at [tex]\( (0, 5) \)[/tex].
- The line [tex]\(y = 2\)[/tex] intersects the [tex]\(x + 2y = 10\)[/tex] line at [tex]\(x = 10 - 2y \Rightarrow x = 10 - 2(2) = 6\)[/tex], giving us the point [tex]\((6, 2)\)[/tex].
The points to be checked are:
- Intersection of [tex]\(x + 2y = 10\)[/tex] and [tex]\(y = 2\)[/tex] [tex]\(\Rightarrow (6, 2)\)[/tex]
- [tex]\(x\)[/tex]-axis and [tex]\(x + 2y = 10\)[/tex] [tex]\(\Rightarrow (10, 0)\)[/tex]
- Intersections with the constraints of non-negativity which are [tex]\((0, 0)\)[/tex] and [tex]\((0, 5)\)[/tex].
5. Evaluate the objective function at these points:
- [tex]\(P(0, 0) = 4(0) + 2(0) = 0\)[/tex]
- [tex]\(P(6, 2) = 4(6) + 2(2) = 24 + 4 = 28\)[/tex]
- [tex]\(P(10, 0) = 4(10) + 2(0) = 40\)[/tex]
- [tex]\(P(0, 5) = 4(0) + 2(5) = 10\)[/tex]
6. Determine the maximum value:
Among these corner points, the maximum value of [tex]\(P\)[/tex] is [tex]\(40\)[/tex], which occurs at the point [tex]\((10, 0)\)[/tex].
Thus, the maximum value of [tex]\(P = 4x + 2y\)[/tex] given the constraints is [tex]\(\boxed{40}\)[/tex].
[tex]\[ \begin{array}{l} x + 2y \leq 10 \\ y \leq 2 \\ x \geq 0 \\ y \geq 0 \end{array} \][/tex]
we need to solve this as a linear programming problem. Let’s follow these steps:
1. Identify the objective function and constraints:
- Objective function: [tex]\(P = 4x + 2y\)[/tex]
- Constraints:
[tex]\[ \begin{array}{l} x + 2y \leq 10 \\ y \leq 2 \\ x \geq 0 \\ y \geq 0 \end{array} \][/tex]
2. Graph the constraints:
- Plot the line [tex]\(x + 2y = 10\)[/tex]. This will be a boundary line.
- Plot the horizontal line [tex]\(y = 2\)[/tex].
- Add the non-negativity constraints [tex]\(x \geq 0\)[/tex] and [tex]\(y \geq 0\)[/tex].
3. Determine the feasible region:
- The feasible region is the area where all constraints overlap and it lies in the first quadrant (both [tex]\(x\)[/tex] and [tex]\(y\)[/tex] are non-negative).
4. Identify the corner points of the feasible region:
- The boundary line [tex]\(x + 2y = 10\)[/tex] intersects the [tex]\(y\)[/tex]-axis at [tex]\( (0, 5) \)[/tex].
- The line [tex]\(y = 2\)[/tex] intersects the [tex]\(x + 2y = 10\)[/tex] line at [tex]\(x = 10 - 2y \Rightarrow x = 10 - 2(2) = 6\)[/tex], giving us the point [tex]\((6, 2)\)[/tex].
The points to be checked are:
- Intersection of [tex]\(x + 2y = 10\)[/tex] and [tex]\(y = 2\)[/tex] [tex]\(\Rightarrow (6, 2)\)[/tex]
- [tex]\(x\)[/tex]-axis and [tex]\(x + 2y = 10\)[/tex] [tex]\(\Rightarrow (10, 0)\)[/tex]
- Intersections with the constraints of non-negativity which are [tex]\((0, 0)\)[/tex] and [tex]\((0, 5)\)[/tex].
5. Evaluate the objective function at these points:
- [tex]\(P(0, 0) = 4(0) + 2(0) = 0\)[/tex]
- [tex]\(P(6, 2) = 4(6) + 2(2) = 24 + 4 = 28\)[/tex]
- [tex]\(P(10, 0) = 4(10) + 2(0) = 40\)[/tex]
- [tex]\(P(0, 5) = 4(0) + 2(5) = 10\)[/tex]
6. Determine the maximum value:
Among these corner points, the maximum value of [tex]\(P\)[/tex] is [tex]\(40\)[/tex], which occurs at the point [tex]\((10, 0)\)[/tex].
Thus, the maximum value of [tex]\(P = 4x + 2y\)[/tex] given the constraints is [tex]\(\boxed{40}\)[/tex].
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