Worksheet on formulating systems of inequalities with real-life applications.

Let’s solve each problem step by step.

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Problem 1: Jenna and her sister saving money

Jenna saves $40 per week.
Her sister saves $30 per week.
They want to save at least $600 total.
Also, they want to save for no more than 20 weeks.

Let:
- x = number of weeks Jenna saves
- y = number of weeks her sister saves

But note: The problem says “each week” — so we assume they both save every week together? Or can they save different numbers of weeks?

Looking at the wording:
> “Jenna was shopping... She saved $40 each week while her sister saved $30 each week. They have decided to save at least $600 for a summer trip and would like to spend no more than 20 weeks.”

It doesn’t say they must save the same number of weeks. So we treat x and y as independent variables.

So:

Total savings: 40x + 30y ≥ 600
Time constraint: x ≤ 20, y ≤ 20 (since neither wants to save more than 20 weeks)
Also, x ≥ 0, y ≥ 0 (can’t save negative weeks)

We are to graph this system and find two possible combinations.

First, simplify the inequality:

40x + 30y ≥ 600
Divide everything by 10:
→ 4x + 3y ≥ 60

Now, let’s find intercepts for the line 4x + 3y = 60:

If x = 0 → 3y = 60 → y = 20
If y = 0 → 4x = 60 → x = 15

So the line goes from (0,20) to (15,0). Since it’s “≥”, we shade above the line.

Also, x ≤ 20, y ≤ 20 — but since our line already hits y=20 at x=0, and x=15 at y=0, and we’re shading above, the feasible region is bounded by:

- Above the line 4x + 3y = 60
- Left of x=20
- Below y=20
- Right of x=0, above y=0

Now, pick two points in that region.

Try x=15, y=0 → 4(15)+3(0)=60 → meets exactly → valid.

Try x=0, y=20 → 4(0)+3(20)=60 → also valid.

Try x=10, y=10 → 4(10)+3(10)=70 ≥ 60 → valid.

Try x=5, y=15 → 4(5)+3(15)=20+45=65 ≥ 60 → valid.

Any of these work.

Two possible combinations:
→ (15, 0): Jenna saves 15 weeks, sister saves 0 weeks → total = 40*15 + 30*0 = 600
→ (0, 20): Jenna saves 0 weeks, sister saves 20 weeks → total = 0 + 600 = 600
→ (10, 10): 400 + 300 = 700 → also works

We’ll pick two easy ones.

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Problem 2: Selling T-shirts and sweatshirts

T-shirt profit: $8 each
Sweatshirt profit: $12 each
Goal: At least $360 profit
Constraint: No more than 40 items total sold
Also, cannot sell negative items.

Let:
- x = number of T-shirts
- y = number of sweatshirts

Inequalities:

Profit: 8x + 12y ≥ 360
Items: x + y ≤ 40
x ≥ 0, y ≥ 0

Simplify profit inequality: divide by 4 → 2x + 3y ≥ 90

Find intercepts for 2x + 3y = 90:

x=0 → 3y=90 → y=30
y=0 → 2x=90 → x=45 → but x+y≤40, so x=45 is not allowed.

Also, x+y=40 → when x=0, y=40; when y=0, x=40.

We need to find where 2x+3y=90 intersects x+y=40.

Solve:

From x+y=40 → x=40−y

Plug into 2x+3y=90:

2(40−y) + 3y = 90
80 − 2y + 3y = 90
80 + y = 90 → y=10 → x=30

So intersection at (30,10)

Feasible region:

Above 2x+3y=90
Below x+y=40
x≥0, y≥0

Vertices of feasible region:

- Intersection point: (30,10)
- Where 2x+3y=90 meets y-axis: (0,30) — check if x+y≤40: 0+30=30 ≤40 → OK
- Where x+y=40 meets x-axis: (40,0) — check profit: 2(40)+3(0)=80 < 90 → NOT in feasible region
- Where x+y=40 meets y-axis: (0,40) — profit: 0 + 120 = 120 ≥ 90 → OK? Wait, 2x+3y= 0 + 120 = 120 ≥ 90 → yes, but wait — our simplified inequality is 2x+3y≥90, which corresponds to original 8x+12y≥360 → 8(0)+12(40)=480≥360 → yes.

Wait — actually, (0,40): profit = 12*40=480 ≥360 → valid. And items=40 ≤40 → valid.

But does (0,40) satisfy 2x+3y≥90? 0 + 120 = 120 ≥90 → yes.

Similarly, (0,30): profit=360 → exact.

And (30,10): profit=8*30 + 12*10=240+120=360 → exact.

What about (40,0)? Profit=320 < 360 → invalid.

So feasible region vertices: (0,30), (30,10), (0,40)? Wait — (0,40) is on x+y=40, and satisfies profit.

But is there a point between (0,30) and (0,40)? Yes, all y from 30 to 40 when x=0.

Actually, the boundary lines:

- From (0,30) along 2x+3y=90 to (30,10)
- Then from (30,10) along x+y=40 to (0,40)? No — because from (30,10) to (0,40) is not straight on either constraint.

Wait — let's think again.

The constraints are:

1. 2x + 3y ≥ 90
2. x + y ≤ 40
3. x ≥ 0, y ≥ 0

The feasible region is bounded by:

- Line A: 2x + 3y = 90 (from (0,30) to (30,10))
- Line B: x + y = 40 (from (30,10) to (0,40))
- Also, from (0,30) up to (0,40) along y-axis? But only if those points satisfy both inequalities.

At x=0:

Constraint 1: 3y ≥ 90 → y ≥ 30
Constraint 2: y ≤ 40
So for x=0, y ∈ [30,40]

Similarly, at y=0:

Constraint 1: 2x ≥ 90 → x ≥ 45
But constraint 2: x ≤ 40 → no solution on y=0 except possibly none.

So the feasible region is a polygon with vertices:

- (0,30)
- (30,10)
- (0,40)

Wait — is (0,40) connected directly to (0,30)? Yes, vertically.

But geometrically, the region is a triangle with vertices at (0,30), (30,10), and (0,40)? That doesn't make sense because (0,40) and (0,30) are on the same vertical line.

Actually, plotting:

- Start at (0,30) — on profit line
- Go along profit line to (30,10)
- Then go along item limit line x