Question

find the maximum value of
p = 7x + 9y
subject to the following constraints:
\

$$\begin{cases} 2x + 8y \\leq 64 \\\\ 3x + 9y \\geq 81 \\\\ x \\geq 0 \\\\ y \\geq 0 \\end{cases}$$

p = ?

Explanation:

Step1: Simplify Constraints

First, simplify the inequalities. For \(2x + 8y \leq 64\), divide by 2: \(x + 4y \leq 32\) or \(x = 32 - 4y\). For \(3x + 9y \geq 81\), divide by 3: \(x + 3y \geq 27\) or \(x = 27 - 3y\). Also, \(x \geq 0\), \(y \geq 0\).

Step2: Find Intersection of Constraints

To find the feasible region, we can find the intersection of \(x + 4y = 32\) and \(x + 3y = 27\). Subtract the second equation from the first: \((x + 4y)-(x + 3y)=32 - 27\) → \(y = 5\). Then substitute \(y = 5\) into \(x + 3y = 27\): \(x + 15 = 27\) → \(x = 12\).

Step3: Analyze Feasible Region Vertices

The feasible region is bounded by the constraints. The vertices are found by intersecting the boundary lines with the axes and with each other.
- Intersection of \(x + 4y = 32\) and \(x = 0\): \(0 + 4y = 32\) → \(y = 8\) (point \((0, 8)\)). But check \(x + 3y \geq 27\): \(0 + 24 = 24 < 27\), so \((0, 8)\) is not in feasible region.
- Intersection of \(x + 3y = 27\) and \(x = 0\): \(0 + 3y = 27\) → \(y = 9\). Check \(x + 4y \leq 32\): \(0 + 36 = 36 > 32\), not feasible.
- Intersection of \(x + 3y = 27\) and \(y = 0\): \(x = 27\). Check \(x + 4y \leq 32\): \(27 \leq 32\), so point \((27, 0)\).
- Intersection of \(x + 4y = 32\) and \(y = 0\): \(x = 32\). Check \(x + 3y \geq 27\): \(32 \geq 27\), so point \((32, 0)\). But wait, earlier intersection of the two main constraints is \((12, 5)\). Wait, maybe better to check the feasible region: the feasible region is where \(x + 3y \geq 27\), \(x + 4y \leq 32\), \(x \geq 0\), \(y \geq 0\). So solve \(x + 3y = 27\) and \(x = 0\): \(y = 9\) (but \(x + 4y = 36 > 32\), so not in \(x + 4y \leq 32\)). Solve \(x + 4y = 32\) and \(y = 0\): \(x = 32\) (and \(x + 3y = 32 \geq 27\), so valid). Solve \(x + 3y = 27\) and \(y = 0\): \(x = 27\) (and \(x + 4y = 27 \leq 32\), valid). And the intersection of \(x + 3y = 27\) and \(x + 4y = 32\) is \((12, 5)\) (as found earlier). Wait, but when \(y = 5\), \(x = 12\), check \(x + 4y = 12 + 20 = 32\) (good), \(x + 3y = 12 + 15 = 27\) (good). Now, check if there are other vertices. Wait, the feasible region is a polygon with vertices at \((27, 0)\), \((32, 0)\), and \((12, 5)\)? Wait no, wait when \(x = 32\), \(y = 0\): \(x + 3y = 32 \geq 27\) (yes), \(x + 4y = 32 \leq 32\) (yes). When \(x = 27\), \(y = 0\): \(x + 4y = 27 \leq 32\) (yes), \(x + 3y = 27 \geq 27\) (yes). When \(x = 12\), \(y = 5\): both constraints satisfied. Wait, but also, is there a vertex at \(y\) - axis? Let's see, for \(x = 0\), \(x + 3y \geq 27\) → \(y \geq 9\), but \(x + 4y \leq 32\) → \(y \leq 8\), which is a contradiction, so no vertex on \(x = 0\) except maybe none. So the feasible region vertices are \((27, 0)\), \((32, 0)\), and \((12, 5)\)? Wait, no, wait when \(x = 32\), \(y = 0\): \(P = 7*32 + 9*0 = 224\). When \(x = 27\), \(y = 0\): \(P = 7*27 + 9*0 = 189\). When \(x = 12\), \(y = 5\): \(P = 7*12 + 9*5 = 84 + 45 = 129\). Wait, but that can't be right, maybe I made a mistake. Wait, no, maybe the feasible region is actually where \(x + 3y \geq 27\) and \(x + 4y \leq 32\), so the region between these two lines and \(x \geq 0\), \(y \geq 0\). Wait, let's graph mentally: \(x + 3y = 27\) is a line with x-intercept 27, y-intercept 9. \(x + 4y = 32\) is a line with x-intercept 32, y-intercept 8. So the feasible region is above \(x + 3y = 27\) and below \(x + 4y = 32\), and \(x \geq 0\), \(y \geq 0\). So the intersection of \(x + 3y = 27\) and \(x + 4y = 32\) is \((12, 5)\) (as before). Then, the other vertices: where \(x + 4y = 32\) meets \(x = 0\): \(y = 8\), but \(x + 3y = 24 < 27\), so not in feasible region. Where \(x + 3y = 27\) meets \(y = 0\): \(x = 27\), and \(x + 4y = 27 \leq 32\), so valid. Where \(x + 4y = 32\) meets \(y = 0\): \(x = 32\), and \(x + 3y = 32 \geq 27\), so valid. So the feasible region has three vertices: \((27, 0)\), \((32, 0)\), and \((12, 5)\)? Wait, but when \(x = 32\), \(y = 0\), \(x + 3y = 32 \geq 27\) (yes), \(x + 4y = 32 \leq 32\) (yes). When \(x = 27\), \(y = 0\), \(x + 4y = 27 \leq 32\) (yes), \(x + 3y = 27 \geq 27\) (yes). When \(x = 12\), \(y = 5\), both constraints satisfied. Now, calculate \(P\) at each vertex:
- At \((27, 0)\): \(P = 7*27 + 9*0 = 189\)
- At \((32, 0)\): \(P = 7*32 + 9*0 = 224\)
- At \((12, 5)\): \(P = 7*12 + 9*5 = 84 + 45 = 129\)

Wait, but is there a mistake here? Wait, the constraint \(x + 3y \geq 27\) and \(x + 4y \leq 32\). Let's check if \((32, 0)\) is in \(x + 3y \geq 27\): \(32 + 0 = 32 \geq 27\), yes. And \(x + 4y = 32 \leq 32\), yes. So that's valid. But wait, when \(y\) increases, \(x\) decreases from \(x + 4y = 32\), but \(x\) must be at least \(27 - 3y\). So when does \(27 - 3y \leq 32 - 4y\)? \(27 - 3y \leq 32 - 4y\) → \(y \leq 5\). So \(y\) can go up to 5, as we found. So the feasible region is from \(y = 0\) to \(y = 5\), with \(x\) between \(27 - 3y\) and \(32 - 4y\). So the vertices are indeed \((27, 0)\), \((32, 0)\), and \((12, 5)\). Wait, but when \(y = 0\), the maximum \(x\) is 32 (from \(x + 4y = 32\)) and minimum \(x\) is 27 (from \(x + 3y = 27\)). So the edge from \((27, 0)\) to \((32, 0)\) is part of the feasible region (since \(x\) between 27 and 32, \(y = 0\) satisfies both constraints). Then, from \((32, 0)\) up along \(x + 4y = 32\) to \((12, 5)\), and from \((12, 5)\) up along \(x + 3y = 27\) to... but \(x + 3y = 27\) at \(y = 9\) is \(x = 0\), but \(x + 4y = 36 > 32\), so that's not feasible. So the feasible region is a triangle with vertices \((27, 0)\), \((32, 0)\), \((12, 5)\). Now, calculating \(P\) at these points, the maximum is at \((32, 0)\) with \(P = 224\)? Wait, but wait, let's check the original constraints again. The original constraint was \(2x + 8y \leq 64\) (so \(x + 4y \leq 32\)) and \(3x + 9y \geq 81\) (so \(x + 3y \geq 27\)). When \(x = 32\), \(y = 0\), \(2x + 8y = 64 \leq 64\) (yes, equality), \(3x + 9y = 96 \geq 81\) (yes). So that's valid. When \(x = 12\), \(y = 5\), \(2x + 8y = 24 + 40 = 64 \leq 64\) (yes), \(3x + 9y = 36 + 45 = 81 \geq 81\) (yes). When \(x = 27\), \(y = 0\), \(2x + 8y = 54 \leq 64\) (yes), \(3x + 9y = 81 \geq 81\) (yes). So all three points are valid. Now, \(P = 7x + 9y\). At \((32, 0)\), \(P = 224\); at \((27, 0)\), \(P = 189\); at \((12, 5)\), \(P = 129\). So the maximum is 224? Wait, but wait, is there a mistake in assuming the feasible region? Wait, maybe I messed up the direction of the inequalities. Let's re-express the constraints:

- \(2x + 8y \leq 64\): below the line \(2x + 8y = 64\)
- \(3x + 9y \geq 81\): above the line \(3x + 9y = 81\)
- \(x \geq 0\), \(y \geq 0\): first quadrant

So the feasible region is the area that is below \(2x + 8y = 64\), above \(3x + 9y = 81\), and in the first quadrant. To find the intersection of \(2x + 8y = 64\) and \(3x + 9y = 81\), we did that earlier: \(x = 12\), \(y = 5\). Now, the intersection of \(3x + 9y = 81\) with \(x = 0\) is \(y = 9\) (point \((0, 9)\)), but check if \((0, 9)\) is below \(2x + 8y = 64\): \(0 + 72 = 72 > 64\), so no. The intersection of \(2x + 8y = 64\) with \(y = 0\) is \(x = 32\) (point \((32, 0)\)), which is above \(3x + 9y = 81\) (since \(3*32 + 0 = 96 > 81\)). The intersection of \(3x + 9y = 81\) with \(y = 0\) is \(x = 27\) (point \((27, 0)\)), which is below \(2x + 8y = 64\) (since \(2*27 + 0 = 54 < 64\)). So the feasible region is a polygon with vertices at \((27, 0)\), \((32, 0)\), and \((12, 5)\) (the intersection of the two lines). Now, when we calculate \(P\) at these points, \((32, 0)\) gives \(P = 7*32 + 9*0 = 224\), \((27, 0)\) gives \(189\), and \((12, 5)\) gives \(129\). So the maximum is 224. Wait, but let's check if there's a mistake in the intersection of the lines. Let's solve \(2x + 8y = 64\) and \(3x + 9y = 81\) again. Multiply first equation by 3: \(6x + 24y = 192\). Multiply second equation by 2: \(6x + 18y = 162\). Subtract: \(6y = 30\) → \(y = 5\). Then \(2x + 40 = 64\) → \(2x = 24\) → \(x = 12\). Correct. So that's right. So the feasible region vertices are correct. Then, the maximum \(P\) is at \((32, 0)\) with \(P = 224\).

Answer:

\(224\)