QuestionSolve the given linear programming problem. Maximize $z=5 x+5 y$ subject to $x \geq 0, y \geq 0, x+y \geq 1,2 x+3 y \leq 12,3 x+2 y \leq 12$
What is the solution? The maximum value of $z$ is $z=$ $\square$ , and it occurs at the point $(x, y)=$ $\square$ . (Type exact answers. Type integers or simplified fractions.)

Studdy Solution

STEP 1

1. We need to maximize the objective function $z = 5x + 5y$ .
2. The constraints are $x \geq 0$ , $y \geq 0$ , $x + y \geq 1$ , $2x + 3y \leq 12$ , and $3x + 2y \leq 12$ .
3. The solution must satisfy all constraints.
4. The solution must be found at one of the vertices of the feasible region.

STEP 2

1. Identify the feasible region.
2. Find the vertices of the feasible region.
3. Evaluate the objective function at each vertex.
4. Determine the maximum value of the objective function.

STEP 3

Identify the feasible region by graphing the constraints.
- The constraint $x \geq 0$ means the solution is in the right half-plane. - The constraint $y \geq 0$ means the solution is in the upper half-plane. - The constraint $x + y \geq 1$ is a line with a slope of -1 passing through (1,0) and (0,1), and the feasible region is above this line. - The constraint $2x + 3y \leq 12$ is a line with a slope of -2/3 passing through (6,0) and (0,4), and the feasible region is below this line. - The constraint $3x + 2y \leq 12$ is a line with a slope of -3/2 passing through (4,0) and (0,6), and the feasible region is below this line.

STEP 4

Find the vertices of the feasible region by solving the system of equations formed by the intersection of the lines:
1. Intersection of $x + y = 1$ and $2x + 3y = 12$ : x + y = 1 \quad \text{(Equation 1)} \] 2x + 3y = 12 \quad \text{(Equation 2)} \]
Solve Equation 1 for $y$ : $y = 1 - x$ .
Substitute into Equation 2: 2x + 3(1 - x) = 12 \] 2x + 3 - 3x = 12 \] -x + 3 = 12 \] x = -9 \quad \text{(Not feasible since } x \geq 0\text{)} \]
2. Intersection of $x + y = 1$ and $3x + 2y = 12$ : x + y = 1 \quad \text{(Equation 1)} \] 3x + 2y = 12 \quad \text{(Equation 3)} \]
Solve Equation 1 for $y$ : $y = 1 - x$ .
Substitute into Equation 3: 3x + 2(1 - x) = 12 \] 3x + 2 - 2x = 12 \] x + 2 = 12 \] x = 10 \quad \text{(Not feasible since } x \leq 4\text{)} \]
3. Intersection of $2x + 3y = 12$ and $3x + 2y = 12$ : 2x + 3y = 12 \quad \text{(Equation 2)} \] 3x + 2y = 12 \quad \text{(Equation 3)} \]
Multiply Equation 2 by 3 and Equation 3 by 2: 6x + 9y = 36 \] 6x + 4y = 24 \]
Subtract the second from the first: 5y = 12 \] y = \frac{12}{5} \]
Substitute back into Equation 2: 2x + 3\left(\frac{12}{5}\right) = 12 \] 2x + \frac{36}{5} = 12 \] 2x = 12 - \frac{36}{5} \] 2x = \frac{60}{5} - \frac{36}{5} \] 2x = \frac{24}{5} \] x = \frac{12}{5} \]
The intersection point is $\left(\frac{12}{5}, \frac{12}{5}\right)$ .
4. Intersection of $x + y = 1$ , $x = 0$ , and $y = 0$ : - The points are $(1,0)$ , $(0,1)$ , and $(0,0)$ .

STEP 5

Evaluate the objective function $z = 5x + 5y$ at each vertex:
1. At $(1, 0)$ : $ z = 5(1) + 5(0) = 5 \]
2. At $(0, 1)$ : $ z = 5(0) + 5(1) = 5 \]
3. At $\left(\frac{12}{5}, \frac{12}{5}\right)$ : $ z = 5\left(\frac{12}{5}\right) + 5\left(\frac{12}{5}\right) = 12 + 12 = 24 \]

STEP 6

Determine the maximum value of the objective function.
The maximum value of $z$ is $24$ , and it occurs at the point $\left(\frac{12}{5}, \frac{12}{5}\right)$ .
The maximum value of $z$ is $z = 24$ , and it occurs at the point $(x, y) = \left(\frac{12}{5}, \frac{12}{5}\right)$ .

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