find the maximum value of\np = 7x + 9y\nsubject to the following constraints:\n\\begin{cases}\n2x + 8y \\leq…

Explanation:

Step1: Simplify the constraints

First, simplify the inequality (2x + 8y\leq64) by dividing both sides by 2, we get (x + 4y\leq32), which can be rewritten as (x\leq32 - 4y). For the inequality (3x + 9y\geq81), divide both sides by 3, we have (x + 3y\geq27), so (x\geq27 - 3y). Also, we know that (x\geq0) and (y\geq0).

Step2: Find the intersection of (x\leq32 - 4y) and (x\geq27 - 3y)

Set (32 - 4y=27 - 3y) (to find the intersection of the two lines (x = 32 - 4y) and (x = 27 - 3y)). Solve for (y): [ \begin{align*} 32-4y&=27 - 3y\ -4y + 3y&=27 - 32\ -y&=- 5\ y&=5 \end{align*} ] Substitute (y = 5) into (x=32 - 4y), we get (x=32-4\times5=32 - 20 = 12). So the intersection point of the two lines is ((12,5)).

Step3: Find the vertices of the feasible region

1. Intersection of (x = 0) and (x + 3y=27): Substitute (x = 0) into (x + 3y=27), we have (3y=27), so (y = 9). The point is ((0,9)). But we need to check if it satisfies (x + 4y\leq32). When (x = 0,y = 9), (x + 4y=0 + 36=36>32), so this point is not in the feasible region.
2. Intersection of (y = 0) and (x + 4y=32): Substitute (y = 0) into (x + 4y=32), we get (x = 32). Check if it satisfies (x + 3y\geq27), when (x = 32,y = 0), (x+3y=32\geq27), so the point is ((32,0)).
3. Intersection of (x = 0) and (y = 0): The point is ((0,0)), but it does not satisfy (x + 3y\geq27) (since (0+0 = 0<27)), so it is not in the feasible region.
4. The intersection of (x + 4y=32) and (x + 3y=27) is ((12,5)) (calculated in step 2).
5. Intersection of (y = 0) and (x + 3y=27): Substitute (y = 0) into (x + 3y=27), we get (x=27). Check if it satisfies (x + 4y\leq32), when (x = 27,y = 0), (x + 4y=27>32) is false, so this point is not in the feasible region.
6. Intersection of (x=32) and (x + 3y=27): If (x = 32), then (32+3y\geq27), (3y\geq - 5), (y\geq-\frac{5}{3}), but since (y\geq0), we check (x = 32,y = 0) which we already considered.

So the feasible region vertices are ((12,5)) and ((32,0)) (we also need to check the intersection of (x + 3y=27) and (x + 4y=32) which we found as ((12,5)) and the intersection of (x + 4y=32) and (y = 0) which is ((32,0)) and we need to check the boundary of (x + 3y=27) and (x\geq0,y\geq0) but the only valid vertices from the intersection of the constraints and non - negativity are ((12,5)) and ((32,0)) and also we can check the intersection of (x + 3y=27) with (y) - axis but it is not in the feasible region, intersection with (x) - axis is not in the feasible region. Also, we can check the point ((0,8)) (since (x + 4y=32) when (y = 8,x = 0)) and check if it satisfies (x + 3y\geq27), when (x = 0,y = 8), (x + 3y=24<27), so it is not in the feasible region.

Step4: Evaluate (P = 7x+9y) at the vertices

1. At the point ((12,5)): (P=7\times12 + 9\times5=84 + 45=129)
2. At the point ((32,0)): (P=7\times32+9\times0 = 224+0 = 224)? Wait, no, we made a mistake. Wait, when (x = 32,y = 0), check the constraint (3x + 9y\geq81), (3\times32+9\times0=96\geq81), and (2x + 8y=64\leq64) (satisfied). But wait, when we consider the constraint (x + 4y\leq32), when (x = 32,y = 0), (32+0 = 32\leq32) (satisfied). But wait, let's check the other vertex. Wait, we missed the intersection of (x + 3y=27) and (y) - axis. When (x = 0), (y = 9), but (x + 4y=0 + 36=36>32), so it is not in the feasible region. Wait, another vertex: intersection of (x + 4y=32) and (y) - axis ((x = 0)): (x = 0,y = 8), check (x + 3y=0+24 = 24<27), not in feasible region. Intersection of (x + 3y=27) and (x) - axis ((y = 0)): (x = 27,y = 0), check (x + 4y=27+0 = 27\leq32), so the point ((27,0)) is also a vertex.

Let's recalculate the vertices:

• Intersection of (x + 3y=27) and (y = 0): (x = 27,y = 0) (since (3x+9\times0\geq81\Rightarrow x\geq27), and (2x + 8\times0\leq64\Rightarrow x\leq32), so (x = 27) is valid as (27\leq32))
• Intersection of (x + 4y=32) and (y = 0): (x = 32,y = 0)
• Intersection of (x + 4y=32) and (x + 3y=27): ((12,5))
• Intersection of (x + 3y=27) and (x = 0): (x = 0,y = 9) (invalid as (x + 4y=36>32))
• Intersection of (x + 4y=32) and (x = 0): (x = 0,y = 8) (invalid as (x + 3y=24<27))

Now re - evaluate (P) at the correct vertices:

1. At ((27,0)): (P=7\times27+9\times0=189)
2. At ((32,0)): (P=7\times32 + 9\times0=224) (but check (3x+9y=96\geq81) and (2x + 8y = 64\leq64), so it is in the feasible region)
3. At ((12,5)): (P=7\times12+9\times5 = 84 + 45=129)

Wait, but we have a mistake in the constraint simplification. Wait, the original constraint (2x + 8y\leq64) is (x + 4y\leq32), and (3x + 9y\geq81) is (x + 3y\geq27). The feasible region is defined by (27 - 3y\leq x\leq32 - 4y), (y\geq0) and (x\geq0).

To find the lower bound of (y), from (x\geq27 - 3y) and (x\geq0), when (x = 0), (27-3y\leq0\Rightarrow y\geq9)? Wait, no, we have (x\geq27 - 3y) and (x\geq0), so (27 - 3y\leq x) and (x\geq0), so (27 - 3y\leq x) and (x\geq0) implies (27 - 3y\leq x) and (x\geq0), so (27-3y\leq x) and (x\geq0), so if (27 - 3y\leq0), then (y\geq9). But when (y\geq9), check (x + 4y\leq32), if (y = 9), (x\leq32-36=- 4), which is not possible since (x\geq0). So there is a contradiction, which means our initial approach to find vertices is wrong.

The correct way is to graph the inequalities:

• (2x + 8y\leq64) (or (x + 4y\leq32)): The line (x + 4y=32) passes through ((32,0)) and ((0,8))
• (3x + 9y\geq81) (or (x + 3y\geq27)): The line (x + 3y=27) passes through ((27,0)) and ((0,9))
• (x\geq0,y\geq0)

The feasible region is the set of points that satisfy (x\geq0,y\geq0), (x + 4y\leq32) and (x + 3y\geq27). To find the feasible region, we need to find the area that is above (x + 3y = 27), below (x + 4y=32) and in the first quadrant ((x\geq0,y\geq0)).

Find the intersection of (x + 4y=32) and (x + 3y=27): As we did before, (x=32 - 4y) and (x=27 - 3y), setting equal: (32-4y=27 - 3y\Rightarrow y = 5), (x = 12). Now, check if this point is in the first quadrant (yes, (x = 12,y = 5\geq0)).

Now, check the intersection of (x + 3y=27) with (x = 0): (y = 9), but when (y = 9), (x + 4y=0 + 36=36>32), so it is not in the feasible region (since it does not satisfy (x + 4y\leq32)).

Intersection of (x + 4y=32) with (y = 0): (x = 32), and when (x = 32,y = 0), (x + 3y=32\geq27), so it is in the feasible region.

Intersection of (x + 3y=27) with (y = 0): (x = 27), and when (x = 27,y = 0), (x + 4y=27\leq32), so it is in the feasible region.

So the feasible region is a polygon with vertices ((27,0)), ((32,0)) and ((12,5))? Wait, no, when (y) increases from 0, the lower bound of (x) is (x=27 - 3y) and the upper bound is (x = 32 - 4y). We need to find the range of (y) for which (27 - 3y\leq32 - 4y) and (27 - 3y\geq0) and (32 - 4y\geq0).

From (27 - 3y\leq32 - 4y), we get (y\leq5) (as before). From (27 - 3y\geq0), (y\leq9). From (32 - 4y\geq0), (y\leq8). So (y\in[0,5]) (since when (y>5), (27 - 3y>32 - 4y) which means there is no (x) that satisfies both (x\geq27 - 3y) and (x\leq32 - 4y)).

So when (y = 0), (x) ranges from (27) to (32) (since (x\geq27) and (x\leq32) when (y = 0)). So the vertices of the feasible region are:

• When (y = 0), the left - most point is ((27[LLM SSE On Failure]