it is 300 tv
and 100 refrigerators
Shipping restrictions. The accompanying graph shows all of the possibilities for the number of refrigerators and the number of TVs that will fit into an 18-wheeler. 300 tv 100 refrigerators
a) Write an inequality to describe this region.
b) Will the truck hold 71 refrigerators and 118 TVs?
c) Will the truck hold 51 refrigerators and 176 TVs?
The Burbank Buy More store is going to make an order which will include at most 60 refrigerators. What is the maximum number of TVs which could also be delivered on the same 18-wheeler? Describe the restrictions this would add to the existing graph.
The next day, the Burbank Buy More decides they will have a television sale so they change their order to include at least 200 TVs. What is the maximum number of refrigerators which could also be delivered in the same truck? Describe the restrictions this would add to the original graph.
first find slope from points (110,0) and (0,330).so slope is -3
equation is y<=-3x+330
so inequality is 3x+y<=330
for part b)
put x=71 and y=118
so we get 331<=330 which is false
so truck can't hold 71 refrigirators and 118 tvs
for part c.
put x=51 and y=176
so we get 329<=330 which is true.so truck hold 51 ref and 176 TV's
To solve these problems, we can create an inequality based on the given information and then plug in the values to check if they satisfy the inequality.
a) The given information states that the truck can hold 300 TVs and 100 refrigerators. To write an inequality for this region, we use the inequality symbol '≤' (less than or equal to) because it includes all the combinations of refrigerators and TVs that can fit in the truck.
So, the inequality can be written as:
300x + 100y ≤ 18
Where 'x' represents the number of refrigerators and 'y' represents the number of TVs.
b) To check if the truck can hold 71 refrigerators and 118 TVs, we substitute these values into the inequality and check if it holds true.
300(71) + 100(118) ≤ 18
21300 + 11800 ≤ 18
33100 ≤ 18
This statement is false, which means the truck cannot hold 71 refrigerators and 118 TVs together.
c) Similarly, for the combination of 51 refrigerators and 176 TVs:
300(51) + 100(176) ≤ 18
15300 + 17600 ≤ 18
32900 ≤ 18
This statement is also false, so the truck cannot hold 51 refrigerators and 176 TVs together.
d) If the Burbank Buy More store wants to order at most 60 refrigerators, we need to find the maximum number of TVs that can also be delivered. This means adding the new restriction that the number of refrigerators must be less than or equal to 60.
So, the modified inequality becomes:
300x + 100y ≤ 18
100y ≤ 18 - 300x
Now, substitute the maximum value of x, which is 60, into the inequality:
100y ≤ 18 - 300(60)
100y ≤ 18 - 18000
100y ≤ -17982
Simplifying further, we get:
y ≤ -179.82
Since the number of TVs cannot be negative, the maximum number of TVs that can be delivered in this case is 0.
e) If the store now wants to order at least 200 TVs, we need to find the maximum number of refrigerators that can also be delivered. This means adding the new restriction that the number of TVs must be greater than or equal to 200.
So, the modified inequality becomes:
300x + 100y ≤ 18
300x ≤ 18 - 100y
Now, substitute the minimum value of y, which is 200, into the inequality:
300x ≤ 18 - 100(200)
300x ≤ 18 - 20000
300x ≤ -19982
Simplifying further, we get:
x ≤ -66.61
Since the number of refrigerators cannot be negative, the maximum number of refrigerators that can be delivered in this case is 0.
Describing the new restrictions on the original graph, for case d) the maximum number of TVs is 0, so the line representing the inequality will be extended horizontally at y = 0. For case e) the maximum number of refrigerators is 0, so the line representing the inequality will be extended vertically at x = 0.