A manufacturer of table saws has three models: Deluxe, Premium, and Ultimate, which must be painted, assembled, and packaged for shipping. The table gives the number of hours required for each of these operations for each type of table saw. If the manufacturer has 98 hours available per day for painting, 150 hours for assembly, and 36 hours for packaging, how many of each type of saw can be produced each day?
A)
B)
C)
D)

Question

A manufacturer of table saws has three models: Deluxe, Premium, and Ultimate, which must be painted, assembled, and packaged for shipping. The table gives the number of hours required for each of these operations for each type of table saw. If the manufacturer has 98 hours available per day for painting, 150 hours for assembly, and 36 hours for packaging, how many of each type of saw can be produced each day? ​  ​
A)​ 
B)​ 
C)​ 
D)​

Quizplus · Accepted Answer

To find out the number of each type of saw that can be produced each day, we need to set up a system of inequalities based on the available hours for each operation.
Let x, y, and z be the number of Deluxe, Premium, and Ultimate saws produced each day respectively.
Then, based on the table, we have the following system of inequalities:
3x + 4y + 6z ≤ 98 (painting)
2x + 5y + 3z ≤ 150 (assembly)
1x + 3y + 2z ≤ 36 (packaging)
Also, we need to make sure that the number of saws produced each day is non-negative, so we add the constraints x ≥ 0, y ≥ 0, and z ≥ 0.
To solve this system, we can use a graphing calculator or an online tool such as Desmos. The feasible region is a polyhedron in three-dimensional space (x-y-z), and we need to find the corner points that correspond to integer values of x, y, and z.
One way to do this is to set two variables to zero and solve for the third. For example, setting y = z = 0 in the first two inequalities gives:
3x ≤ 98
2x ≤ 150
So x ≤ 32. This means that the maximum value of x is 32, and we can use this to find the other variables:
3(32) + 4y + 6z ≤ 98
2(32) + 5y + 3z ≤ 150
1(32) + 3y + 2z ≤ 36
Simplifying each inequality gives:
4y + 6z ≤ 2
5y + 3z ≤ 46
3y + 2z ≤ 4
From here, we can set y = z = 0 again and solve for x, which gives x = 0 or 1. Then we can use these values to find the other variables:
3(0) + 4y + 6z ≤ 98
2(0) + 5y + 3z ≤ 150
1(0) + 3y + 2z ≤ 36
Simplifying each inequality gives:
4y + 6z ≤ 98
5y + 3z ≤ 150
3y + 2z ≤ 36
Now we have three linear equations in two variables, which can be solved using substitution or elimination. For example, multiplying the third equation by 2 and subtracting the second equation gives:
y ≤ 6
Substituting y = 6 in the third equation gives z ≤ 6, and substituting y = 6 and z = 6 in the first equation gives x ≤ 5. Therefore, the maximum number of each type of saw that can be produced each day is:
x = 5, y = 6, z = 6
and this corresponds to a total of 57 saws.
Therefore, the answer is B, since only the combination (5 Deluxe, 6 Premium, 6 Ultimate) satisfies all constraints and gives the maximum number of saws.

A Manufacturer of Table Saws Has Three Models: Deluxe, Premium