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Maximize [tex]$F=6x + 5y$[/tex] subject to:
[tex]\[ x + y \leq 4 \][/tex]
[tex]\[ 2x + y \leq 7 \][/tex]
[tex]\[ x \geq 0 \][/tex]
[tex]\[ y \geq 0 \][/tex]

a. Maximum value

b. Point where maximum value occurs


Sagot :

To maximize the function [tex]\( F = 6x + 5y \)[/tex] subject to the constraints:

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

we proceed with the following steps:

### Step 1: Identify the Feasible Region
Firstly, let's identify the feasible region determined by the constraints:

- The line [tex]\( x + y = 4 \)[/tex] can be rewritten as [tex]\( y = 4 - x \)[/tex]. This represents a line that passes through the points [tex]\( (0,4) \)[/tex] and [tex]\( (4,0) \)[/tex].
- The line [tex]\( 2x + y = 7 \)[/tex] can be rewritten as [tex]\( y = 7 - 2x \)[/tex]. This represents a line that passes through the points [tex]\( (0,7) \)[/tex] and [tex]\( (3.5,0) \)[/tex].

Next, determine the intersection points and which portion of the plane satisfies all the constraints [tex]\( x \geq 0 \)[/tex] and [tex]\( y \geq 0 \)[/tex]. The region of interest is bounded by these lines and the coordinate axes.

### Step 2: Find the Intersection Points
Calculate the intersection points of the lines:

1. Intersection of [tex]\( x + y = 4 \)[/tex] and [tex]\( 2x + y = 7 \)[/tex]:

[tex]\[ \begin{cases} x + y = 4 \\ 2x + y = 7 \end{cases} \][/tex]

Subtract the first equation from the second:

[tex]\[ (2x + y) - (x + y) = 7 - 4 \\ x = 3 \][/tex]

Substitute [tex]\( x = 3 \)[/tex] back into [tex]\( x + y = 4 \)[/tex]:

[tex]\[ 3 + y = 4 \\ y = 1 \][/tex]

So, the intersection point is [tex]\( (3, 1) \)[/tex].

### Step 3: Evaluate the Objective Function at Corner Points
Evaluate [tex]\( F = 6x + 5y \)[/tex] at each vertex of the feasible region. The vertices are the intersection points and the points where the lines intersect the axes:

1. At [tex]\( (0,0) \)[/tex]:
[tex]\[ F = 6(0) + 5(0) = 0 \][/tex]

2. At [tex]\( (4,0) \)[/tex]:
[tex]\[ F = 6(4) + 5(0) = 24 \][/tex]

3. At [tex]\( (0,7) \)[/tex]:
[tex]\[ F = 6(0) + 5(7) = 35 \quad (\text{Note: This point does not satisfy } x + y \leq 4) \][/tex]

4. At [tex]\( (3,1) \)[/tex]:
[tex]\[ F = 6(3) + 5(1) = 18 + 5 = 23 \][/tex]

### Step 4: Determine the Maximum Value
Compare the values obtained at each feasible vertex:

- [tex]\( F(0, 0) = 0 \)[/tex]
- [tex]\( F(4, 0) = 24 \)[/tex]
- [tex]\( F(3, 1) = 23 \)[/tex]

The maximum value is [tex]\( F = 24 \)[/tex] achieved at the point [tex]\( (3, 1) \)[/tex].

### Conclusion

[a] The maximum value of [tex]\( F \)[/tex] is [tex]\(\boxed{23.0}\)[/tex].

[b] The point where this maximum value occurs is [tex]\(\boxed{(3.0, 1.0)}\)[/tex].

Therefore, the final answer is:
(a) The maximum value [tex]\(F\)[/tex] is [tex]\(23.0\)[/tex].
(b) The point where the maximum value occurs is [tex]\((3.0, 1.0)\)[/tex].

These are the optimal solutions within the feasible region for the given linear programming problem.
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