Math Problem Statement

A ceramies 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 9 hours to fire. A platter takes 14 hours to prepare, 3 hours to paint, and 4 hours to fire. A bread basket takes 4 hours to prepare, 14 hours to paint, and 7 hours to fire. If the workshop has 92 hours for prep time, 70 hours for painting, and 109 hours for firing, how many of each can be made?

Solution

This problem can be solved by setting up a system of linear equations based on the hours required for each activity (preparing, painting, and firing) and the total hours available for each.

Let:

  • xx = number of serving bowls
  • yy = number of platters
  • zz = number of bread baskets

We can now express the constraints based on the available hours for preparation, painting, and firing.

Preparation time:

  • A serving bowl takes 3 hours to prepare, so 3x3x hours total for xx serving bowls.
  • A platter takes 14 hours to prepare, so 14y14y hours total for yy platters.
  • A bread basket takes 4 hours to prepare, so 4z4z hours total for zz bread baskets.
  • The workshop has 92 hours available for preparation. 3x+14y+4z=923x + 14y + 4z = 92

Painting time:

  • A serving bowl takes 2 hours to paint, so 2x2x hours total for xx serving bowls.
  • A platter takes 3 hours to paint, so 3y3y hours total for yy platters.
  • A bread basket takes 14 hours to paint, so 14z14z hours total for zz bread baskets.
  • The workshop has 70 hours available for painting. 2x+3y+14z=702x + 3y + 14z = 70

Firing time:

  • A serving bowl takes 9 hours to fire, so 9x9x hours total for xx serving bowls.
  • A platter takes 4 hours to fire, so 4y4y hours total for yy platters.
  • A bread basket takes 7 hours to fire, so 7z7z hours total for zz bread baskets.
  • The workshop has 109 hours available for firing. 9x+4y+7z=1099x + 4y + 7z = 109

We now have the following system of equations:

1. & \quad 3x + 14y + 4z = 92 \\ 2. & \quad 2x + 3y + 14z = 70 \\ 3. & \quad 9x + 4y + 7z = 109$$ I'll solve this system of equations to find the values of $$x$$, $$y$$, and $$z$$.The workshop can make: - 8 serving bowls ($$x = 8$$), - 4 platters ($$y = 4$$), - 3 bread baskets ($$z = 3$$). Would you like more details on how these values were derived or any further clarifications? Here are five related questions to expand on this problem: 1. How would the solution change if the workshop had more hours available for preparation? 2. What if the preparation time for each bread basket increased by 2 hours? 3. Can the workshop make more items if the firing time per platter was reduced to 2 hours? 4. How many total items can the workshop produce given these time constraints? 5. What is the maximum number of platters the workshop can make if no bowls or bread baskets are made? **Tip:** In systems of linear equations, altering one constraint often affects the feasibility or quantity of solutions, so recalculating consistently is key to maintaining accuracy.

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Math Problem Analysis

Mathematical Concepts

Algebra
Systems of Linear Equations

Formulas

3x + 14y + 4z = 92 (preparation time)
2x + 3y + 14z = 70 (painting time)
9x + 4y + 7z = 109 (firing time)

Theorems

Solving Systems of Linear Equations using Substitution or Elimination

Suitable Grade Level

Grades 10-12