Questions: Minimize z=3x+5y 2y+7x ≥ 20 7y+2x ≥ 20 Subject to y+x ≥ 5 x ≥ 0 y ≥ 0 Minimum is at x= y=

Transcript text: Minimize $\quad z=3 x+5 y$ $2 y+7 x \geq 20$ $7 y+2 x \geq 20$ Subject to $y+x \geq 5$ $x \geq 0$ $y \geq 0$ Minimum is $\square$ at $x=$ $\square$ $y=$ $\square$ Question Help: Message instructor

Solution

Solution Steps

To solve this linear programming problem, we need to find the values of $x$ and $y$ that minimize the objective function $z = 3x + 5y$ while satisfying all the given constraints. We can use the scipy.optimize.linprog function from the SciPy library in Python to achieve this.

Step 1: Identify the Constraints and Objective Function

We are given the following linear programming problem:

Objective function: \[ z = 3x + 5y \]

Subject to the constraints: \[ \begin{align*}

& \quad 2y + 7x \geq 20 \\
& \quad 7y + 2x \geq 20 \\
& \quad y + x \geq 5 \\
& \quad x \geq 0 \\
& \quad y \geq 0 \end{align*} \]

Step 2: Convert Inequalities to Equalities

To find the feasible region, we convert the inequalities to equalities and find the intersection points:

$ 2y + 7x = 20 $
$ 7y + 2x = 20 $
$ y + x = 5 $

Step 3: Find Intersection Points

Solve the system of equations to find the intersection points.

Intersection of $ 2y + 7x = 20 $ and $ 7y + 2x = 20 $:

\[ \begin{align_} 2y + 7x &= 20 \quad \text{(1)} \\ 7y + 2x &= 20 \quad \text{(2)} \end{align_} \]

Multiply (1) by 7 and (2) by 2 to eliminate $ y $: \[ \begin{align_} 14y + 49x &= 140 \quad \text{(3)} \\ 14y + 4x &= 40 \quad \text{(4)} \end{align_} \]

Subtract (4) from (3): \[ 45x = 100 \implies x = \frac{100}{45} = \frac{20}{9} \]

Substitute $ x = \frac{20}{9} $ into (1): \[ 2y + 7 \left( \frac{20}{9} \right) = 20 \implies 2y + \frac{140}{9} = 20 \implies 2y = 20 - \frac{140}{9} = \frac{180}{9} - \frac{140}{9} = \frac{40}{9} \implies y = \frac{20}{9} \]

So, the intersection point is: \[ \left( \frac{20}{9}, \frac{20}{9} \right) \]

Intersection of $ 2y + 7x = 20 $ and $ y + x = 5 $:

\[ \begin{align_} 2y + 7x &= 20 \quad \text{(1)} \\ y + x &= 5 \quad \text{(5)} \end{align_} \]

Substitute $ y = 5 - x $ into (1): \[ 2(5 - x) + 7x = 20 \implies 10 - 2x + 7x = 20 \implies 5x = 10 \implies x = 2 \]

Substitute $ x = 2 $ into (5): \[ y = 5 - 2 = 3 \]

So, the intersection point is: \[ (2, 3) \]

Intersection of $ 7y + 2x = 20 $ and $ y + x = 5 $:

\[ \begin{align_} 7y + 2x &= 20 \quad \text{(2)} \\ y + x &= 5 \quad \text{(5)} \end{align_} \]

Substitute $ y = 5 - x $ into (2): \[ 7(5 - x) + 2x = 20 \implies 35 - 7x + 2x = 20 \implies -5x = -15 \implies x = 3 \]

Substitute $ x = 3 $ into (5): \[ y = 5 - 3 = 2 \]

So, the intersection point is: \[ (3, 2) \]

Step 4: Evaluate the Objective Function at Each Vertex

Evaluate $ z = 3x + 5y $ at each intersection point:

At $ \left( \frac{20}{9}, \frac{20}{9} \right) $: \[ z = 3 \left( \frac{20}{9} \right) + 5 \left( \frac{20}{9} \right) = \frac{60}{9} + \frac{100}{9} = \frac{160}{9} \approx 17.7778 \]
At $ (2, 3) $: \[ z = 3(2) + 5(3) = 6 + 15 = 21 \]
At $ (3, 2) $: \[ z = 3(3) + 5(2) = 9 + 10 = 19 \]