Linear Programming Calculator

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Example

Created on 2024-06-20Asked by Harper Carter (Solvelet student)

Maximize

z = 3x + 4y

subject to the constraints:

\begin{cases} x + 2y \leq 8 \\ 3x + y \leq 9 \\ x \geq 0, \, y \geq 0 \end{cases}

Solution

To solve the linear programming problem, we first identify the feasible region defined by the constraints. The constraints are:

x + 2y \leq 8 \quad \text{(1)}

3x + y \leq 9 \quad \text{(2)}

x \geq 0, \, y \geq 0 \quad \text{(3)}

We graph these inequalities to find the feasible region. Next, we determine the corner points of the feasible region by solving the system of equations formed by the intersection of the boundary lines: 1. Intersection of

x + 2y = 8

and

3x + y = 9

x + 2y = 8 \quad (i)

3x + y = 9 \quad (ii)

Multiply equation (ii) by 2:

6x + 2y = 18

Subtract equation (i) from this result:

(6x + 2y) - (x + 2y) = 18 - 8

5x = 10

x = 2

Substitute

x = 2

back into equation (i):

2 + 2y = 8

2y = 6

y = 3

So, the intersection point is

(2, 3)

. 2. Intersection of

x + 2y = 8

and

x = 0

x = 0 \Rightarrow 2y = 8 \Rightarrow y = 4

So, the intersection point is

(0, 4)

. 3. Intersection of

3x + y = 9

and

y = 0

3x = 9 \Rightarrow x = 3

So, the intersection point is

(3, 0)

. 4. Intersection of

x = 0

and

y = 0

(0, 0)

We evaluate

z = 3x + 4y

at these points:

\begin{aligned} &\text{At } (0, 0): z = 3(0) + 4(0) = 0 \\ &\text{At } (0, 4): z = 3(0) + 4(4) = 16 \\ &\text{At } (3, 0): z = 3(3) + 4(0) = 9 \\ &\text{At } (2, 3): z = 3(2) + 4(3) = 6 + 12 = 18 \end{aligned}

The maximum value of

z = 18

occurs at

(2, 3)

. Therefore, the solution is

x = 2

y = 3

, and the maximum value of

z

is 18. Solved on Solvelet with Basic AI Model

Some of the related questions asked by Mateo Taylor on Solvelet

1. Maximize the objective function

z = 3x + 2y

subject to the constraints:

x + y \leq 10

2x + y \leq 15

x \geq 0

, and

y \geq 0

.2. Minimize the objective function

z = 5x + 4y

subject to the constraints:

x + y \geq 6

2x + 3y \geq 12

x \geq 0

, and

y \geq 0

DefinitionLines and planes in space are three-dimensional geometric objects. Vector DIR is the direction of a line, P is the point on the line OR a point to determine a plane, and the VPLANE is the normal of the plane. Equations give us the location of the lines and planes as well as the direction. Such as: The line through (1,2,3) having direction vector (4,5,6) is expressed as r(t)=(1,2,3)+t(4,5,6) vector equation 4 The equation to the plane passing through (1,2,3) with its normal vector: (4,5,6) is 4(x−1)+5(y−2)+6(z−3)=0

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