To optimize the equation [tex]\( P = 4x + 3y \)[/tex] subject to the constraints [tex]\( x \geq 0 \)[/tex], [tex]\( x \leq 4 \)[/tex], [tex]\( y \geq -1 \)[/tex], and [tex]\( y \leq 5 \)[/tex], we need to find the maximum value of the objective function within the feasible region defined by these constraints.
1. Identify the constraints and plot the feasible region:
- [tex]\( x \geq 0 \)[/tex]: This is the region to the right of the y-axis.
- [tex]\( x \leq 4 \)[/tex]: This is the region to the left of the line [tex]\( x = 4 \)[/tex].
- [tex]\( y \geq -1 \)[/tex]: This is the region above the line [tex]\( y = -1 \)[/tex].
- [tex]\( y \leq 5 \)[/tex]: This is the region below the line [tex]\( y = 5 \)[/tex].
Combining these constraints, the feasible region is a rectangle with the vertices at:
[tex]\[
(0, -1), (0, 5), (4, -1), (4, 5)
\][/tex]
2. Calculate the value of the objective function [tex]\( P \)[/tex] at each vertex of the feasible region:
- At [tex]\( (0, -1) \)[/tex]:
[tex]\[
P = 4(0) + 3(-1) = -3
\][/tex]
- At [tex]\( (0, 5) \)[/tex]:
[tex]\[
P = 4(0) + 3(5) = 15
\][/tex]
- At [tex]\( (4, -1) \)[/tex]:
[tex]\[
P = 4(4) + 3(-1) = 16 - 3 = 13
\][/tex]
- At [tex]\( (4, 5) \)[/tex]:
[tex]\[
P = 4(4) + 3(5) = 16 + 15 = 31
\][/tex]
3. Determine the maximum value:
- The values of [tex]\( P \)[/tex] at the vertices are: -3, 15, 13, and 31.
- The maximum value among these is 31.
Therefore, the maximum value of [tex]\( P = 4x + 3y \)[/tex] within the given constraints is [tex]\( \boxed{31} \)[/tex].
