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 \(7x + 7y \leq 49\) (y=0)

Set \(y = 0\) in \(7x + 7y = 49\), we get \(7x = 49\), so \(x=\frac{49}{7}=7\)? Wait, no, wait the other line? Wait, the green box is for x when y=0. Wait, the constraints are \(8x + 6y \leq 48\) and \(7x + 7y \leq 49\), \(x\geq0\), \(y\geq0\). When y=0, for \(8x + 6(0) \leq 48\), \(8x \leq 48\), \(x \leq 6\). For \(7x + 7(0) \leq 49\), \(7x \leq 49\), \(x \leq 7\). But the feasible region's x-intercept (y=0) is determined by the more restrictive constraint. Wait, the graph shows x-intercepts at 6 (blue) and 7 (red), but the feasible region is the intersection. Wait, no, the green box is for the x when y=0, looking at the table, the second row is y=0, so we need to find x when y=0, considering the constraints. Wait, the first constraint when y=0: \(8x \leq 48\) → \(x \leq 6\). The second constraint when y=0: \(7x \leq 49\) → \(x \leq 7\). But the feasible region is the intersection, so the x-intercept (y=0) is at x=6? Wait, no, the graph has blue line at x=6 (when y=0, 8x=48 → x=6) and red line at x=7 (7x=49 → x=7). But the feasible region is the area that satisfies both constraints. So when y=0, x must satisfy both \(8x \leq 48\) and \(7x \leq 49\). The more restrictive is \(8x \leq 48\) → x ≤6. Wait, but the table's second row is y=0, so x is? Wait, maybe I misread. Wait, the problem's table: first row (0,0), second row (?, 0), third row (0,?), fourth row (?,?). Let's find the corner points.

First, find intersection of \(8x + 6y = 48\) and \(7x + 7y = 49\). Let's solve the system:

From \(7x + 7y = 49\), divide by 7: \(x + y = 7\) → \(y = 7 - x\).

Substitute into \(8x + 6y = 48\):

\(8x + 6(7 - x) = 48\)

\(8x + 42 - 6x = 48\)

\(2x + 42 = 48\)

\(2x = 6\) → \(x = 3\), then \(y = 7 - 3 = 4\).

Now, corner points:

1. (0,0): intersection of x=0 and y=0.

1. (x, 0): intersection of y=0 and the feasible region. The constraints when y=0: \(8x \leq 48\) → x ≤6; \(7x \leq 49\) → x ≤7. So the x-intercept is at x=6 (since 6 <7, so the feasible region's x-intercept is (6,0)).

1. (0, y): intersection of x=0 and feasible region. For x=0, \(6y \leq 48\) → y ≤8; \(7y \leq 49\) → y ≤7. So y=7 (more restrictive), so (0,7).

1. Intersection of \(8x + 6y = 48\) and \(7x + 7y = 49\), which we found as (3,4).

Wait, but the table's second row is y=0, so x is 6? Wait, no, let's check the first constraint when y=0: 8x =48 → x=6. So the second row (y=0) has x=6.

Wait, maybe I made a mistake earlier. Let's re-express:

To find x when y=0 (second row of the table), we use the constraint that, when y=0, the maximum x is from \(8x + 6(0) \leq 48\) → \(x = 48/8 = 6\). Because the other constraint \(7x + 7(0) \leq 49\) gives x=7, but the feasible region is the set of points that satisfy all constraints, so x must be ≤6 (from 8x+6y≤48) and ≤7 (from 7x+7y≤49), so x≤6. Thus, when y=0, x=6.

So the second row's x is 6.

Step1: Solve for x when y=0

Set \(y = 0\) in \(8x + 6y = 48\) (since it's the constraint that gives the x-intercept for the feasible region's boundary at y=0, as it's more restrictive than \(7x + 7y = 49\) at y=0).
\(8x + 6(0) = 48\)
\(8x = 48\)
\(x = \frac{48}{8} = 6\)

Answer:

6