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Charlotte is knitting hats and scarves to sell on her online store. She has 20 balls of yarn and needs one-half a ball for each hat and a full ball for each scarf.

It takes her 5 hours to make one hat and 4 hours to make one scarf, and she can spend a maximum of 110 hours knitting this month. Charlotte makes a profit of [tex]$\$[/tex]8[tex]$ for every hat she sells and $[/tex]\[tex]$6$[/tex] for every scarf she sells.

The system of inequalities that represents this situation is:

[tex]\[
\left\{
\begin{array}{l}
\frac{1}{2} x + y \leq 20 \\
5x + 4y \leq 110
\end{array}
\right.
\][/tex]

This system is shown graphed in the first quadrant with the number of hats she makes along the [tex]$x$[/tex]-axis and the number of scarves she makes along the [tex]$y$[/tex]-axis.

Sagot :

GinnyAnswer
Sure, let's analyze the given problem step-by-step.

### Step 1: Understand the Constraints and Variables
Let's define our variables:
- Let [tex]\( x \)[/tex] be the number of hats Charlotte makes.
- Let [tex]\( y \)[/tex] be the number of scarves Charlotte makes.

### Step 2: Set Up the Inequalities
We are given two constraints:

1. Yarn Constraint:
- Each hat requires [tex]\( \frac{1}{2} \)[/tex] ball of yarn.
- Each scarf requires 1 ball of yarn.
- Charlotte has a total of 20 balls of yarn.

This gives us the inequality:
[tex]\[ \frac{1}{2}x + y \leq 20 \][/tex]

2. Time Constraint:
- Each hat takes 5 hours to knit.
- Each scarf takes 4 hours to knit.
- Charlotte can spend a maximum of 110 hours knitting.

This gives us the inequality:
[tex]\[ 5x + 4y \leq 110 \][/tex]

### Step 3: Graph the Inequalities
Graphically, these inequalities represent the feasible region for [tex]\( x \)[/tex] and [tex]\( y \)[/tex].

### Step 4: Find the Intersection Points (Vertices)
First, let's solve the system of equations formed by converting the inequalities into equalities:

[tex]\[ \begin{cases} \frac{1}{2} x + y = 20 \\ 5x + 4y = 110 \end{cases} \][/tex]

#### Solving for [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:

1. From [tex]\(\frac{1}{2}x + y = 20\)[/tex]:
[tex]\[ y = 20 - \frac{1}{2}x \][/tex]

2. Substitute [tex]\( y \)[/tex] into [tex]\( 5x + 4y = 110 \)[/tex]:
[tex]\[ 5x + 4\left(20 - \frac{1}{2}x\right) = 110 \][/tex]
[tex]\[ 5x + 80 - 2x = 110 \][/tex]
[tex]\[ 3x = 30 \][/tex]
[tex]\[ x = 10 \][/tex]

3. Substitute [tex]\( x = 10 \)[/tex] back into [tex]\( y = 20 - \frac{1}{2}x \)[/tex]:
[tex]\[ y = 20 - \frac{1}{2}(10) = 20 - 5 = 15 \][/tex]

So, the intersection point is [tex]\( (10, 15) \)[/tex].

### Step 5: Compute the Profit
Charlotte's profit function is:
[tex]\[ Profit = 8x + 6y \][/tex]

Substitute [tex]\( x = 10 \)[/tex] and [tex]\( y = 15 \)[/tex]:
[tex]\[ Profit = 8(10) + 6(15) = 80 + 90 = 170 \][/tex]

### Step 6: Conclusion
From our calculations, the optimal combination is making 10 hats and 15 scarves, yielding a maximum profit of \[tex]$170. So, Charlotte should make 10 hats and 15 scarves to achieve her maximum possible profit of \$[/tex]170.
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