Answer :
To find the maximum value of [tex]\( P = 4x + 5y \)[/tex] subject to the constraints:
[tex]\[ \begin{cases} x + 3y \leq 13 \\ 3x + 2y \leq 25 \\ x \geq 0 \\ y \geq 0 \end{cases} \][/tex]
we follow these steps:
1. Identify the constraint lines:
- [tex]\( x + 3y = 13 \)[/tex]
- [tex]\( 3x + 2y = 25 \)[/tex]
2. Graph the inequality constraints:
- [tex]\( x + 3y \leq 13 \)[/tex]
- [tex]\( 3x + 2y \leq 25 \)[/tex]
3. Determine the feasible region:
Plot the lines and find the region that satisfies all the constraints simultaneously, including [tex]\( x \geq 0 \)[/tex] and [tex]\( y \geq 0 \)[/tex].
4. Find the vertices of the feasible region:
Calculate the points of intersection of the lines, including where they intersect the axes.
- Intersection of [tex]\( x + 3y = 13 \)[/tex] and [tex]\( x \)[/tex]-axis: Set [tex]\( y = 0 \)[/tex], solve [tex]\( x = 13 \)[/tex]. Point [tex]\( (13, 0) \)[/tex].
- Intersection of [tex]\( x + 3y = 13 \)[/tex] and [tex]\( y \)[/tex]-axis: Set [tex]\( x = 0 \)[/tex], solve [tex]\( 3y = 13 \)[/tex], [tex]\( y = \frac{13}{3} \approx 4.33 \)[/tex]. Point [tex]\( (0, \frac{13}{3}) \)[/tex].
- Intersection of [tex]\( 3x + 2y = 25 \)[/tex] and [tex]\( x \)[/tex]-axis: Set [tex]\( y = 0 \)[/tex], solve [tex]\( 3x = 25 \)[/tex], [tex]\( x \approx 8.33 \)[/tex]. Point [tex]\( ( \frac{25}{3}, 0) \approx (8.33, 0) \)[/tex].
- Intersection of [tex]\( 3x + 2y = 25 \)[/tex] and [tex]\( y \)[/tex]-axis: Set [tex]\( x = 0 \)[/tex], solve [tex]\( 2y = 25 \)[/tex], [tex]\( y = 12.5 \)[/tex]. Point [tex]\( (0, 12.5) \)[/tex].
- Intersection of [tex]\( x + 3y = 13 \)[/tex] and [tex]\( 3x + 2y = 25 \)[/tex]:
[tex]\[ \begin{cases} x + 3y = 13 \\ 3x + 2y = 25 \end{cases} \][/tex]
Solve the system of equations:
[tex]\[ \text{Multiply the first equation by 3: } 3x + 9y = 39 \][/tex]
[tex]\[ \text{Subtract the second equation from the first: } (3x + 9y) - (3x + 2y) = 39 - 25 \][/tex]
[tex]\[ 7y = 14 \implies y = 2 \][/tex]
Substituting [tex]\( y = 2 \)[/tex] into the first equation:
[tex]\[ x + 3(2) = 13 \implies x + 6 = 13 \implies x = 7 \][/tex]
Intersection point is [tex]\( (7, 2) \)[/tex].
5. Evaluate [tex]\( P = 4x + 5y \)[/tex] at each vertex:
- At [tex]\( (13, 0) \)[/tex]: [tex]\( P = 4(13) + 5(0) = 52 \)[/tex]
- At [tex]\( (0, \frac{13}{3}) \)[/tex]: [tex]\( P = 4(0) + 5(\frac{13}{3}) \approx 21.67 \)[/tex]
- At [tex]\( (\frac{25}{3}, 0) \)[/tex]: [tex]\( P = 4(\frac{25}{3}) + 5(0) \approx 33.33 \)[/tex]
- At [tex]\( (0, 12.5) \)[/tex]: [tex]\( P = 4(0) + 5(12.5) = 62.5 \)[/tex]
- At [tex]\( (7, 2) \)[/tex]: [tex]\( P = 4(7) + 5(2) = 28 + 10 = 38 \)[/tex]
6. Select the maximum value:
The maximum value of [tex]\( P \)[/tex] among these points occurs at [tex]\( (7, 2) \)[/tex] and is:
[tex]\[ P = 38 \][/tex]
So, the maximum value of [tex]\( P = 4x + 5y \)[/tex] subject to the given constraints is [tex]\( \boxed{38} \)[/tex].
[tex]\[ \begin{cases} x + 3y \leq 13 \\ 3x + 2y \leq 25 \\ x \geq 0 \\ y \geq 0 \end{cases} \][/tex]
we follow these steps:
1. Identify the constraint lines:
- [tex]\( x + 3y = 13 \)[/tex]
- [tex]\( 3x + 2y = 25 \)[/tex]
2. Graph the inequality constraints:
- [tex]\( x + 3y \leq 13 \)[/tex]
- [tex]\( 3x + 2y \leq 25 \)[/tex]
3. Determine the feasible region:
Plot the lines and find the region that satisfies all the constraints simultaneously, including [tex]\( x \geq 0 \)[/tex] and [tex]\( y \geq 0 \)[/tex].
4. Find the vertices of the feasible region:
Calculate the points of intersection of the lines, including where they intersect the axes.
- Intersection of [tex]\( x + 3y = 13 \)[/tex] and [tex]\( x \)[/tex]-axis: Set [tex]\( y = 0 \)[/tex], solve [tex]\( x = 13 \)[/tex]. Point [tex]\( (13, 0) \)[/tex].
- Intersection of [tex]\( x + 3y = 13 \)[/tex] and [tex]\( y \)[/tex]-axis: Set [tex]\( x = 0 \)[/tex], solve [tex]\( 3y = 13 \)[/tex], [tex]\( y = \frac{13}{3} \approx 4.33 \)[/tex]. Point [tex]\( (0, \frac{13}{3}) \)[/tex].
- Intersection of [tex]\( 3x + 2y = 25 \)[/tex] and [tex]\( x \)[/tex]-axis: Set [tex]\( y = 0 \)[/tex], solve [tex]\( 3x = 25 \)[/tex], [tex]\( x \approx 8.33 \)[/tex]. Point [tex]\( ( \frac{25}{3}, 0) \approx (8.33, 0) \)[/tex].
- Intersection of [tex]\( 3x + 2y = 25 \)[/tex] and [tex]\( y \)[/tex]-axis: Set [tex]\( x = 0 \)[/tex], solve [tex]\( 2y = 25 \)[/tex], [tex]\( y = 12.5 \)[/tex]. Point [tex]\( (0, 12.5) \)[/tex].
- Intersection of [tex]\( x + 3y = 13 \)[/tex] and [tex]\( 3x + 2y = 25 \)[/tex]:
[tex]\[ \begin{cases} x + 3y = 13 \\ 3x + 2y = 25 \end{cases} \][/tex]
Solve the system of equations:
[tex]\[ \text{Multiply the first equation by 3: } 3x + 9y = 39 \][/tex]
[tex]\[ \text{Subtract the second equation from the first: } (3x + 9y) - (3x + 2y) = 39 - 25 \][/tex]
[tex]\[ 7y = 14 \implies y = 2 \][/tex]
Substituting [tex]\( y = 2 \)[/tex] into the first equation:
[tex]\[ x + 3(2) = 13 \implies x + 6 = 13 \implies x = 7 \][/tex]
Intersection point is [tex]\( (7, 2) \)[/tex].
5. Evaluate [tex]\( P = 4x + 5y \)[/tex] at each vertex:
- At [tex]\( (13, 0) \)[/tex]: [tex]\( P = 4(13) + 5(0) = 52 \)[/tex]
- At [tex]\( (0, \frac{13}{3}) \)[/tex]: [tex]\( P = 4(0) + 5(\frac{13}{3}) \approx 21.67 \)[/tex]
- At [tex]\( (\frac{25}{3}, 0) \)[/tex]: [tex]\( P = 4(\frac{25}{3}) + 5(0) \approx 33.33 \)[/tex]
- At [tex]\( (0, 12.5) \)[/tex]: [tex]\( P = 4(0) + 5(12.5) = 62.5 \)[/tex]
- At [tex]\( (7, 2) \)[/tex]: [tex]\( P = 4(7) + 5(2) = 28 + 10 = 38 \)[/tex]
6. Select the maximum value:
The maximum value of [tex]\( P \)[/tex] among these points occurs at [tex]\( (7, 2) \)[/tex] and is:
[tex]\[ P = 38 \][/tex]
So, the maximum value of [tex]\( P = 4x + 5y \)[/tex] subject to the given constraints is [tex]\( \boxed{38} \)[/tex].
