Solve the problem.
-A ceramics workshop makes serving bowls, platters, and bread baskets to sell at its Winter Festival. A serving bowl takes 3 hours to prepare, 2 hours to paint, and 10 hours to fire. A platter takes 17 hours to prepare, 3 hours
To paint, and 4 hours to fire. A bread basket takes 4 hours to prepare, 17 hours to paint, and 7 hours to fire. If the
Workshop has 94 hours for prep time, 58 hours for painting, and 90 hours for firing, how many of each can be
Made?
A) 7 serving bowls, 5 platters, 3 bread baskets
B) 4 serving bowls, 2 platters, 6 bread baskets
C) 6 serving bowls, 4 platters, 2 bread baskets
D) 2 serving bowls, 6 platters, 4 bread baskets

Question

Quizplus · Accepted Answer

To find the answer, we need to create a system of equations and solve for the number of each item that can be made based on the given time constraints. Let x, y, and z represent the number of serving bowls, platters, and bread baskets made, respectively. Then, we have the following system of equations:
3x + 17y + 4z ≤ 94 (prep time)
2x + 3y + 17z ≤ 58 (painting time)
10x + 4y + 7z ≤ 90 (firing time)
We want to maximize the total number of items made, which is x + y + z. To solve for x, y, and z, we can use substitution or elimination. Using elimination, we can multiply the first equation by 4, the second equation by 17, and the third equation by 2 and add them together:
12x + 68y + 16z ≤ 376
34x + 51y + 289z ≤ 986
20x + 8y + 14z ≤ 180
Then, we can multiply the third equation by 17 and subtract it from the second equation:
14x + 47y + 231z ≤ 478
Subtracting twice the first equation from this, we get:
41y + 167z ≤ 290
Finally, we can see that the only integer solution to this inequality is y = 4 and z = 2. Substituting these values back into the third equation, we get x = 6. Therefore, the answer is C: 6 serving bowls, 4 platters, and 2 bread baskets can be made in the given time.

Solve the Problem. -A Ceramics Workshop Makes Serving Bowls, Platters, and Bread Baskets