Let x be the number of packages of hot dogs and y be the number of packages of buns. Begin by finding the minimum number of hot dogs and buns in terms of packages. Make an organized table and write a system of inequalities that represents the situation.
B
b
The points in the shaded region represent the possible purchases.
A
a
Graph:
B
b
Possible Purchases
4 packages of hot dogs, 5 packages of buns
5 packages of hot dogs, 6 packages of buns
6 packages of hot dogs, 5 packages of buns
Practice makes perfect
a
Let x be the number of packages of hot dogs and y be the number of packages of buns. With this, we will write three inequalities that represent the situation. Let's first write an inequality for the number of packages making an organized table.
Packages of hot dogs
Packages of Buns
Verbal Expression
Algebraic Expression
Verbal Expression
Algebraic Expression
Minimum number of hot dogs
40
Minimum number of buns
40
Number of hot dogs in one package
10
Number of buns in one package
8
Minimum number of packages of hot dogs
40/10= 4
Minimum number of packages of buns
40/8= 5
Number of packages of hot dogs is at least 4.
x≥ 4
Number of packages of buns is at least 5.
y≥ 5
Next, we will write an inequality for the cost proceeding in the same way.
Verbal Expression
Algebraic Expression
Cost of x packages of hot dogs ($)
3.5 x
Cost of y packages of buns ($)
2.5 y
Total cost is less than or equal to $35.
3.5 x+ 2.5 y≤ 35
Now we have three inequalities to write a system.
x ≥ 4 & (I) y ≥ 5 & (II) 3.5x+2.5y ≤ 35 & (III)
Let's begin by drawing the first inequality. We will first find the boundary line by replacing the inequality sign with the equals sign.
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Inequality I &Boundary Line I
x ≥ 4 &x = 4
The boundary line of Inequality I is a vertical line that passes through the point (4,0). Since the number of packages cannot be negative the line will be bound by the axes. It will also be solid because the inequality is non-strict. Let's draw it!
The inequality says that the points with the x-coordinates greater than or equal to 4 are solutions. Therefore, we will shade region to the right of the boundary line.
We will graph the second inequality thinking in the same way.
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Inequality II &Boundary Line II
y ≥ 5 & y = 5
The boundary line will be a horizontal line that passes through the point (0,5). We will shade the region above the line because the points with the y-coordinates greater than or equal to 5 are solutions to the inequality.
Finally, we will graph the last inequality.
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Inequality III &Boundary Line
3.5x+2.5y ≤ 35 & 3.5x+2.5y = 35
To graph the boundary line, we will find its intercepts. We will substitute y= 0 for the x-intercept and x= 0 for the y-intercept.
3.5x+2.5y = 35
Operation
x-intercept
y-intercept
Substitution
3.5x+2.5( 0) = 35
3.5( 0)+2.5y = 35
Calculation
x=10
y=14
Point
(10,0)
(0,14)
Now that we found intercepts, we can plot them and graph the line.
Finally, we will test an arbitrary point to decide which side we should shade. To make the math bit easier, we can test the point (0,0).
