To find the maximum value of [tex]\( P = 9x + 8y \)[/tex] subject to the constraints [tex]\( 8x + 6y \leq 48 \)[/tex], [tex]\( 7x + 7y \leq 49 \)[/tex], [tex]\( x \geq 0 \)[/tex], and [tex]\( y \geq 0 \)[/tex], we need to identify the feasible region determined by these constraints and then calculate [tex]\( P \)[/tex] at the corner points of this feasible region.
First, let's list the constraints:
1. [tex]\( 8x + 6y \leq 48 \)[/tex]
2. [tex]\( 7x + 7y \leq 49 \)[/tex]
3. [tex]\( x \geq 0 \)[/tex]
4. [tex]\( y \geq 0 \)[/tex]
Here's the process to find the corner points of the feasible region:
1. Identify [tex]\( x \)[/tex]- and [tex]\( y \)[/tex]-intercepts of the boundaries defined by the inequalities:
- For [tex]\( 8x + 6y \leq 48 \)[/tex]:
- When [tex]\( x = 0 \)[/tex]: [tex]\( 6y = 48 \)[/tex] → [tex]\( y = 8 \)[/tex]
- When [tex]\( y = 0 \)[/tex]: [tex]\( 8x = 48 \)[/tex] → [tex]\( x = 6 \)[/tex]
- For [tex]\( 7x + 7y \leq 49 \)[/tex]:
- When [tex]\( x = 0 \)[/tex]: [tex]\( 7y = 49 \)[/tex] → [tex]\( y = 7 \)[/tex]
- When [tex]\( y = 0 \)[/tex]: [tex]\( 7x = 49 \)[/tex] → [tex]\( x = 7 \)[/tex]
2. Solve for intersection points of the lines [tex]\( 8x + 6y = 48 \)[/tex] and [tex]\( 7x + 7y = 49 \)[/tex]:
- Convert the equations to find the intersection point:
- [tex]\( 8x + 6y = 48 \)[/tex]
- [tex]\( 7x + 7y = 49 \)[/tex]
- Solving these two equations simultaneously:
Multiply the second equation by [tex]\( \frac{6}{7} \)[/tex] to align coefficients of [tex]\( y \)[/tex]:
- [tex]\( 8x + 6y = 48 \)[/tex]
- [tex]\( 6x + 6y = 42 \)[/tex]
Subtract the second equation from the first:
- [tex]\( (8x + 6y) - (6x + 6y) = 48 - 42 \)[/tex]
- [tex]\( 2x = 6 \)[/tex] → [tex]\( x = 3 \)[/tex]
Substitute [tex]\( x = 3 \)[/tex] back into [tex]\( 7x + 7y = 49 \)[/tex]:
- [tex]\( 7(3) + 7y = 49 \)[/tex] → [tex]\( 21 + 7y = 49 \)[/tex] → [tex]\( 7y = 28 \)[/tex] → [tex]\( y = 4 \)[/tex]
Now, the corner points are:
- [tex]\( (0, 0) \)[/tex]
- [tex]\( (6, 0) \)[/tex]
- [tex]\( (0, 7) \)[/tex]
- [tex]\( (3, 4) \)[/tex]
3. Calculate [tex]\( P \)[/tex] at each corner point:
- At [tex]\( (0, 0) \)[/tex]: [tex]\( P = 9(0) + 8(0) = 0 \)[/tex]
- At [tex]\( (6, 0) \)[/tex]: [tex]\( P = 9(6) + 8(0) = 54 \)[/tex]
- At [tex]\( (0, 7) \)[/tex]: [tex]\( P = 9(0) + 8(7) = 56 \)[/tex]
- At [tex]\( (3, 4) \)[/tex]: [tex]\( P = 9(3) + 8(4) = 27 + 32 = 59 \)[/tex]
4. Determine the maximum [tex]\( P \)[/tex]:
- The values of [tex]\( P \)[/tex] at the corner points are 0, 54, 56, and 59.
- The maximum value is [tex]\( 59 \)[/tex].
Therefore, the profit [tex]\( P \)[/tex] at each corner point and the maximum value of [tex]\( P \)[/tex] are:
[tex]\[
\begin{array}{ccc}
x & y & P \\
0 & 0 & 0 \\
6 & 0 & 54 \\
0 & 7 & 56 \\
3 & 4 & 59 \\
\end{array}
\][/tex]
Thus, the maximum profit [tex]\( P \)[/tex] is 59, which occurs at the corner point [tex]\( (3, 4) \)[/tex].
