Math Problem Statement

A company makes two types of biscuits: Jumbo and Regular. The oven can cook at most 300 biscuits per day. Each jumbo biscuit requires 2 oz of flour, each regular biscuit requires 1 oz of flour, and there is 400 oz of flour available. The income from each jumbo biscuit is $0.12 and from each regular biscuit is $0.07. How many of each size biscuit should be made to maximize income? What is the maximum income?

Solution

This is an optimization problem that can be solved using linear programming. Let's break it down:

Problem Summary:

  • The company makes two types of biscuits: Jumbo and Regular.
  • The oven can bake up to 300 biscuits per day.
  • Flour constraints: 400oz400 \, \text{oz} of flour available.
  • Flour usage:
    • Jumbo biscuit: 2oz2 \, \text{oz} per biscuit.
    • Regular biscuit: 1oz1 \, \text{oz} per biscuit.
  • Profit:
    • Jumbo biscuit: $0.12\$0.12 per biscuit.
    • Regular biscuit: $0.07\$0.07 per biscuit.

Let:

  • xx = Number of Jumbo biscuits.
  • yy = Number of Regular biscuits.

Objective Function:

Maximize profit PP: P=0.12x+0.07yP = 0.12x + 0.07y

Constraints:

  1. Oven capacity: x+y300x + y \leq 300
  2. Flour usage: 2x+y4002x + y \leq 400
  3. Non-negativity: x0x \geq 0, y0y \geq 0

Solving the System:

We'll solve these linear inequalities to find the values of xx and yy that maximize the profit.

  1. Graphical Method: The feasible region will be a polygon on the coordinate plane.
  2. Corner Points: We find the profit at each corner point and choose the one that gives the maximum profit.

Step 1: Find the intersection points of the constraints

  • x+y=300x + y = 300
  • 2x+y=4002x + y = 400

We can solve these equations by substitution or elimination to find the intersection points:

Intersection Point:

  • Subtract x+y=300x + y = 300 from 2x+y=4002x + y = 400: x=100x = 100 y=200y = 200

Step 2: Evaluate the objective function at the feasible points

  • (0,300)(0, 300)
  • (200,0)(200, 0)
  • (100,200)(100, 200)

Calculate Profit PP:

  1. P(0,300)=0.12(0)+0.07(300)=$21.00P(0, 300) = 0.12(0) + 0.07(300) = \$21.00
  2. P(200,0)=0.12(200)+0.07(0)=$24.00P(200, 0) = 0.12(200) + 0.07(0) = \$24.00
  3. P(100,200)=0.12(100)+0.07(200)=$26.00P(100, 200) = 0.12(100) + 0.07(200) = \$26.00

Conclusion:

The company should make 100 Jumbo biscuits and 200 Regular biscuits to maximize profit, and the maximum income will be $26.00.

Would you like more details, or do you have any other questions?

Related Questions:

  1. What would happen if the flour available was reduced to 350 oz?
  2. How would the solution change if the profit per Jumbo biscuit increased to $0.15?
  3. Can this problem be solved using the Simplex Method instead of the graphical method?
  4. How would the solution change if the oven capacity increased to 350 biscuits?
  5. What if the flour required for Jumbo biscuits decreased to 1.5 oz per biscuit?

Tip:

Always check the feasibility of each corner point when solving linear programming problems. It ensures the best possible solution within the constraints.

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Math Problem Analysis

Mathematical Concepts

Linear Programming
Optimization
Inequalities
Profit Maximization

Formulas

P = 0.12x + 0.07y
x + y <= 300
2x + y <= 400
x >= 0, y >= 0

Theorems

Linear Programming Corner Point Theorem

Suitable Grade Level

Grades 10-12