Question

find the maximum value of
p = 9x + 8y
subject to the following constraints:
begin by finding the coordinates of the corner points.
\\(\

$$\begin{cases}8x + 6y \\leq 48\\\\7x + 7y \\leq 49\\\\x \\geq 0\\\\y \\geq 0\\end{cases}$$

\\)

Explanation:

Step1: Find x-intercept of \(8x + 6y \leq 48\) (y=0)

Set \(y = 0\) in \(8x + 6y = 48\), so \(8x = 48\), then \(x=\frac{48}{8}=6\). But wait, also check \(7x + 7y \leq 49\) when \(y = 0\): \(7x=49\Rightarrow x = 7\). But the feasible region is intersection, so for the corner point with \(y = 0\), we take the minimum x from the two constraints? Wait, no, the corner points are where the boundaries intersect or meet axes. Wait, the first corner point with \(y = 0\): let's solve \(8x+6(0)=48\Rightarrow x = 6\), and \(7x+7(0)=49\Rightarrow x = 7\). But the inequality \(8x + 6y \leq 48\) and \(7x + 7y \leq 49\), when \(y = 0\), the more restrictive is \(8x \leq 48\) (x=6) or \(7x \leq 49\) (x=7)? Wait, no, the feasible region is the intersection of all constraints. So \(x \geq 0\), \(y \geq 0\), \(8x + 6y \leq 48\), \(7x + 7y \leq 49\). So when \(y = 0\), \(8x \leq 48\Rightarrow x \leq 6\), and \(7x \leq 49\Rightarrow x \leq 7\). So the intersection is \(x \leq 6\) (since 6 < 7). Wait, but the graph shows x-intercepts at 6 (blue line) and 7 (red line). Wait, maybe the corner points are:

1. (0,0)
2. Intersection of \(8x + 6y = 48\) and \(x = 0\): \(6y = 48\Rightarrow y = 8\), but check \(7x + 7y = 49\) when \(x=0\): \(7y = 49\Rightarrow y = 7\). So intersection of \(x=0\) with the more restrictive: \(y \leq 7\) (from \(7x + 7y \leq 49\)) and \(y \leq 8\) (from \(8x + 6y \leq 48\)), so \(y = 7\) (0,7).

3. Intersection of \(7x + 7y = 49\) (simplify to \(x + y = 7\)) and \(8x + 6y = 48\). Let's solve \(x + y = 7\) (so \(y = 7 - x\)) substitute into \(8x + 6(7 - x)=48\Rightarrow 8x + 42 - 6x = 48\Rightarrow 2x = 6\Rightarrow x = 3\), then \(y = 4\). So (3,4).

4. Intersection of \(8x + 6y = 48\) and \(y = 0\): (6,0)

5. Intersection of \(7x + 7y = 49\) and \(y = 0\): (7,0) – but wait, is (7,0) in \(8x + 6y \leq 48\)? \(8(7)+0=56>48\), so (7,0) is not in the feasible region. So the corner points are (0,0), (0,7), (3,4), (6,0).

Now, let's list all corner points:

- (0,0): \(P = 9(0) + 8(0) = 0\)
- (0,7): \(P = 9(0) + 8(7) = 56\)
- (3,4): \(P = 9(3) + 8(4) = 27 + 32 = 59\)
- (6,0): \(P = 9(6) + 8(0) = 54\)

Wait, but wait, let's confirm the intersection of \(8x + 6y = 48\) and \(7x + 7y = 49\). Solve \(x + y = 7\) (divided 7x+7y=49 by 7) and \(8x + 6y = 48\). Multiply \(x + y = 7\) by 6: \(6x + 6y = 42\). Subtract from \(8x + 6y = 48\): \(2x = 6\Rightarrow x = 3\), then \(y = 7 - 3 = 4\). Correct.

Now, the corner point with \(y = 0\): when \(y = 0\), the feasible x is from \(8x \leq 48\) (x=6) because \(7x \leq 49\) (x=7) is less restrictive? Wait, no, when \(y = 0\), \(8x + 0 \leq 48\Rightarrow x \leq 6\), and \(7x + 0 \leq 49\Rightarrow x \leq 7\). So the maximum x with \(y = 0\) is 6 (since 6 ≤ 7), so (6,0) is a corner point.

Now, calculate P at each corner:

- (0,0): 0
- (0,7): 56
- (3,4): 27 + 32 = 59
- (6,0): 54

So the maximum P is 59. Wait, but let's check the initial table: the first corner point after (0,0) is (?, 0). Wait, maybe the question first is to find the x-coordinate when y=0. Let's solve for the corner point with y=0: from the constraints, when y=0, the feasible x is the minimum of x from 8x ≤ 48 (x=6) and 7x ≤ 49 (x=7), so x=6? Wait, no, 8x + 6y ≤ 48 when y=0: x ≤ 6; 7x + 7y ≤ 49 when y=0: x ≤ 7. So the intersection is x ≤ 6, so the corner point is (6, 0). So the green box is x=6.

Wait, the problem first asks to find the coordinates of the corner points, starting with (0,0), then (?, 0), then (0,?), then (?,?). So for (?, 0): solve 8x + 6(0) = 48 (since 7x + 7(0)=49 gives x=7, but 8x+6y=48 at y=0 gives x=6, and since 8x+6y ≤48 is a constraint, so the corner point on x-axis is (6, 0). So the green box is 6.

Then, for (0,?): solve 7x + 7y = 49 (since 8x + 6y=48 at x=0 gives y=8, but 7x+7y=49 at x=0 gives y=7, so the corner point on y-axis is (0,7).

Then the intersection point (3,4) as solved.

Now, to find maximum P=9x+8y:

- (0,0): 0
- (6,0): 9*6 + 8*0 = 54
- (0,7): 9*0 + 8*7 = 56
- (3,4): 9*3 + 8*4 = 27 + 32 = 59

So maximum value is 59.

But let's go step by step as per the problem: first, find the x when y=0.

Step1: Find x when y=0

Set \(y = 0\) in the constraint \(8x + 6y \leq 48\) (since it's a boundary for the feasible region's x-axis intercept, and check with \(7x + 7y \leq 49\)).

For \(8x + 6(0) = 48\):
\(8x = 48\)
\(x = \frac{48}{8} = 6\)

For \(7x + 7(0) = 49\):
\(7x = 49\)
\(x = 7\)

But the feasible region is the intersection, so the x-intercept for the feasible region is the minimum of 6 and 7, which is 6 (since \(8x + 6y \leq 48\) is more restrictive here, as 6 < 7). So the corner point is (6, 0), so the x-coordinate is 6.

Then, to find the maximum P:

Calculate P at each corner point:

1. (0, 0): \(P = 9(0) + 8(0) = 0\)
2. (6, 0): \(P = 9(6) + 8(0) = 54\)
3. (0, 7): Solve for y when x=0: from \(7x + 7y = 49\) (more restrictive than \(8x + 6y = 48\) at x=0: \(6y = 48\Rightarrow y=8\), but \(7y = 49\Rightarrow y=7\), so y=7). \(P = 9(0) + 8(7) = 56\)
4. (3, 4): Intersection of \(8x + 6y = 48\) and \(7x + 7y = 49\). Solve:
- From \(7x + 7y = 49\), divide by 7: \(x + y = 7\Rightarrow y = 7 - x\)
- Substitute into \(8x + 6y = 48\): \(8x + 6(7 - x) = 48\)
- \(8x + 42 - 6x = 48\)
- \(2x = 6\Rightarrow x = 3\), then \(y = 7 - 3 = 4\)
- \(P = 9(3) + 8(4) = 27 + 32 = 59\)

So the maximum value of P is 59.

Answer:

The maximum value of \(P\) is \(\boxed{59}\) (and the x-coordinate for \(y=0\) is 6).