How Do I solve this system of linear equalities word problem?
"Sandy makes $2 profit on every cup of lemonade that she sells and $1 on every cupcake that she sells. Sandy wants to sell at least 5 cups of lemonade and at least 5 cupcakes per day. She wants to earn at least $25 per day. Show and describe all the possible combinations of lemonade and cupcakes that Sandy needs to sell to meet her goals. List two possible combinations."
number of cups of lemonade --- x
number of cupcakes ---- y
2x + y ≥ 25, where x > 5, y > 5
x .. y
5 15
6 13
7 11
...
Let
x = # cups of lemonade
y = # cupcakes
What have they told you?
x >= 5
y >= 5
2x + 1y >= 25
So, start with the smallest value of x, and then let it grow.
Start with the smallest acceptable profit: 25
x y profit
5 15 25
6 13 25
...
Now, raise the profit to 26 and do it again
5 16 26
6 14 26
...
Or, graph all three of those lines, and shade the solution sets.
Then just pick any points in the shaded area.
C = cupcakes
p = profit
Sandy makes $2 profit on every cup of lemonade that she sells and $1 on every cupcake that she sells neans:
p = $2 L + $1C
p = 2 L + C
She wants to earn at least $25 per day means:
p ≥ 25
2 L + C ≥ 25
Sandy wants to sell at least 5 cups of lemonade and at least 5 cupcakes per day means:
L ≥ 5
C ≥ 5
Start with L = 5
2 L + C ≥ 25
2 ∙ 5 + C ≥ 25
10 + C ≥ 25
C ≥ 15
L = 6
2 L + C ≥ 25
2 ∙ 6 + C ≥ 25
12 + C ≥ 25
C ≥ 13
L = 7
2 L + C ≥ 25
2 ∙ 7 + C ≥ 25
14 + C ≥ 25
C ≥ 11
L = 8
2 L + C ≥ 25
2 ∙ 8 + C ≥ 25
16 + C ≥ 25
C ≥ 9
L = 9
2 L + C ≥ 25
2 ∙ 9 + C ≥ 25
18 + C ≥ 25
C ≥ 7
L = 10
2 L + C ≥ 25
2 ∙ 10 + C ≥ 25
20 + C ≥ 25
C ≥ 5
L = 11
2 L + C ≥ 25
2 ∙ 11 + C ≥ 25
22 + C ≥ 25
C ≥ 3
L = 12
2 L + C ≥ 25
2 ∙ 12 + C ≥ 25
24 + C ≥ 25
C ≥ 1
For L = 13
2 L + C ≥ 25
2 ∙ 13 + C ≥ 25
26 + C ≥ 25
C ≥ - 1
For L ≥ 13 , C is negative so you must reject cobinations L ≥ 13 C. In other words, for L ≥ 13 the lemonade sold would earn more than $25, so she wouldn't even sell cupcakes.
So possible combinations are:
L = 5 , C ≥ 15
L = 6 , C ≥ 13
L = 7 , C ≥ 11
L = 8 , C ≥ 9
L = 9 , C ≥ 7
L = 10 , C ≥ 5
L = 11 , C ≥ 3
L = 12 , C ≥ 1
This is the MINIMUM number of combinations that earns the required profit.
Of course she can sell more than this.
To solve this system of linear equalities word problem, we can set up a system of equations based on the given information:
Let:
x = the number of cups of lemonade Sandy sells per day
y = the number of cupcakes Sandy sells per day
From the problem, we have the following information:
1) Sandy makes $2 profit on every cup of lemonade sold:
Profit from lemonade = 2x
2) Sandy makes $1 profit on every cupcake sold:
Profit from cupcakes = y
3) Sandy wants to sell at least 5 cups of lemonade per day:
x ≥ 5
4) Sandy wants to sell at least 5 cupcakes per day:
y ≥ 5
5) Sandy wants to earn at least $25 per day:
Profit from lemonade + Profit from cupcakes ≥ 25 OR 2x + y ≥ 25
We can now solve this system of linear equalities by substituting the constraints into the equations.
To find the possible combinations, we can plug in different values for x and y that satisfy all the given conditions.
Let's list two possible combinations:
Combination 1:
If Sandy sells 5 cups of lemonade (x = 5) and 15 cupcakes (y = 15), the profit would be:
Profit from lemonade = 2x = 2 * 5 = $10
Profit from cupcakes = y = 15 * $1 = $15
Total profit = Profit from lemonade + Profit from cupcakes = $10 + $15 = $25
This combination satisfies all the given conditions.
Combination 2:
If Sandy sells 6 cups of lemonade (x = 6) and 4 cupcakes (y = 4), the profit would be:
Profit from lemonade = 2x = 2 * 6 = $12
Profit from cupcakes = y = 4 * $1 = $4
Total profit = Profit from lemonade + Profit from cupcakes = $12 + $4 = $16
This combination does not meet the condition of earning at least $25 per day.
Hence, one possible combination is selling 5 cups of lemonade and 15 cupcakes, which would result in earning at least $25 per day.