Question

1. mary knits scarves and sweaters to sell. scarves take 2 hours to knit and sweaters take 10 hours. mary would like to spend no more than 40 hours per week knitting and knit at least 5 items per week.

a. write and graph a system of linear inequalities:
b. write two possible solutions:
i.
ii.

Explanation:

Part a: Writing the System of Linear Inequalities

Let \( x \) be the number of scarves and \( y \) be the number of sweaters.

Step 1: Time Constraint

Scarves take 2 hours, sweaters take 10 hours, and total time ≤ 40 hours:
\( 2x + 10y \leq 40 \) (simplify: \( x + 5y \leq 20 \))

Step 2: Item Constraint

At least 5 items per week:
\( x + y \geq 5 \)

Step 3: Non - negativity (since number of items can't be negative)

\( x \geq 0 \), \( y \geq 0 \)

Graphing (Brief Description)

1. For \( x + 5y \leq 20 \):

• Find intercepts: When \( x = 0 \), \( y = 4 \); when \( y = 0 \), \( x = 20 \). Draw a solid line, shade below (since \( \leq \)).

1. For \( x + y \geq 5 \):

• Intercepts: \( x = 5 \) (when \( y = 0 \)), \( y = 5 \) (when \( x = 0 \)). Draw a solid line, shade above (since \( \geq \)).

1. For \( x \geq 0 \) and \( y \geq 0 \), shade in the first quadrant. The feasible region is where all shadings overlap.

Part b: Possible Solutions

A solution \((x, y)\) must satisfy \( x + 5y \leq 20 \), \( x + y \geq 5 \), \( x \geq 0 \), \( y \geq 0 \).

i. Solution 1:

Let \( x = 5 \), \( y = 1 \)

• Check time: \( 2(5)+10(1)=10 + 10 = 20\leq40 \)
• Check items: \( 5 + 1 = 6\geq5 \)

So \((5, 1)\) is a solution (5 scarves, 1 sweater).

ii. Solution 2:

Let \( x = 0 \), \( y = 4 \)

• Check time: \( 2(0)+10(4)=40\leq40 \)
• Check items: \( 0 + 4 = 4\)? Wait, no. Wait, \( x + y\geq5 \), so \( x = 5 \), \( y = 0 \):
• Time: \( 2(5)+10(0)=10\leq40 \)
• Items: \( 5+0 = 5\geq5 \)

Or \( x = 3 \), \( y = 2 \):

• Time: \( 2(3)+10(2)=6 + 20 = 26\leq40 \)
• Items: \( 3 + 2 = 5\geq5 \)

Let's use \((3, 2)\): 3 scarves, 2 sweaters.

Final Answers (for part b)

i. \(\boldsymbol{(5, 1)}\) (5 scarves, 1 sweater)
ii. \(\boldsymbol{(3, 2)}\) (3 scarves, 2 sweaters)

(Note: Other valid solutions like \((10, 0)\) (10 scarves, 0 sweaters) or \((0, 4)\) is invalid for items, so correct ones must satisfy \( x + y\geq5 \).)

Answer:

Part a: Writing the System of Linear Inequalities

Let \( x \) be the number of scarves and \( y \) be the number of sweaters.

Step 1: Time Constraint

Scarves take 2 hours, sweaters take 10 hours, and total time ≤ 40 hours:
\( 2x + 10y \leq 40 \) (simplify: \( x + 5y \leq 20 \))

Step 2: Item Constraint

At least 5 items per week:
\( x + y \geq 5 \)

Step 3: Non - negativity (since number of items can't be negative)

\( x \geq 0 \), \( y \geq 0 \)

Graphing (Brief Description)

1. For \( x + 5y \leq 20 \):

• Find intercepts: When \( x = 0 \), \( y = 4 \); when \( y = 0 \), \( x = 20 \). Draw a solid line, shade below (since \( \leq \)).

1. For \( x + y \geq 5 \):

• Intercepts: \( x = 5 \) (when \( y = 0 \)), \( y = 5 \) (when \( x = 0 \)). Draw a solid line, shade above (since \( \geq \)).

1. For \( x \geq 0 \) and \( y \geq 0 \), shade in the first quadrant. The feasible region is where all shadings overlap.

Part b: Possible Solutions

A solution \((x, y)\) must satisfy \( x + 5y \leq 20 \), \( x + y \geq 5 \), \( x \geq 0 \), \( y \geq 0 \).

i. Solution 1:

Let \( x = 5 \), \( y = 1 \)

• Check time: \( 2(5)+10(1)=10 + 10 = 20\leq40 \)
• Check items: \( 5 + 1 = 6\geq5 \)

So \((5, 1)\) is a solution (5 scarves, 1 sweater).

ii. Solution 2:

Let \( x = 0 \), \( y = 4 \)

• Check time: \( 2(0)+10(4)=40\leq40 \)
• Check items: \( 0 + 4 = 4\)? Wait, no. Wait, \( x + y\geq5 \), so \( x = 5 \), \( y = 0 \):
• Time: \( 2(5)+10(0)=10\leq40 \)
• Items: \( 5+0 = 5\geq5 \)

Or \( x = 3 \), \( y = 2 \):

• Time: \( 2(3)+10(2)=6 + 20 = 26\leq40 \)
• Items: \( 3 + 2 = 5\geq5 \)

Let's use \((3, 2)\): 3 scarves, 2 sweaters.

Final Answers (for part b)

i. \(\boldsymbol{(5, 1)}\) (5 scarves, 1 sweater)
ii. \(\boldsymbol{(3, 2)}\) (3 scarves, 2 sweaters)

(Note: Other valid solutions like \((10, 0)\) (10 scarves, 0 sweaters) or \((0, 4)\) is invalid for items, so correct ones must satisfy \( x + y\geq5 \).)