Math Problem Statement
The Phony Bologna Meat Company offers two platters. Platter A comes with 4 hamburgers, 3 hot dogs, and 2 pig's feet, and costs $15. Platter B comes with 3 hamburgers, 4 hot dogs, and 5 pig's feet, and costs $12.
A picnic organizer requires 220 hamburgers, 270 hot dogs, and 250 pig's feet. (There can be leftovers, but these are the minimum requirements.) What is the minimum cost (in dollars)?
Solution
To determine the minimum cost to meet the picnic organizer's requirements using platters A and B, we first define the variables:
- : number of Platter A
- : number of Platter B
The constraints based on the picnic organizer's requirements are:
- (hamburgers)
- (hot dogs)
- (pig's feet)
The objective function to minimize is the total cost:
Step 1: Set up the constraints
We have the following system of inequalities:
4x + 3y \geq 220 \\ 3x + 4y \geq 270 \\ 2x + 5y \geq 250 \end{cases}$$ ### Step 2: Solve for $$y$$ in terms of $$x$$ 1. From $$4x + 3y \geq 220$$: $$y \geq \frac{220 - 4x}{3}$$ 2. From $$3x + 4y \geq 270$$: $$y \geq \frac{270 - 3x}{4}$$ 3. From $$2x + 5y \geq 250$$: $$y \geq \frac{250 - 2x}{5}$$ ### Step 3: Determine the feasible region To find the feasible region, plot the inequalities on a graph and find the intersection points. We solve these inequalities by equating two of the constraints at a time. 1. **Intersection of** $$4x + 3y = 220$$ **and** $$3x + 4y = 270$$: Multiply the first equation by 4 and the second by 3: $$16x + 12y = 880$$ $$9x + 12y = 810$$ Subtract the second from the first: $$7x = 70$$ $$x = 10$$ Substitute $$x = 10$$ into $$4x + 3y = 220$$: $$40 + 3y = 220$$ $$3y = 180$$ $$y = 60$$ Point: $$(10, 60)$$ 2. **Intersection of** $$4x + 3y = 220$$ **and** $$2x + 5y = 250$$: Multiply the first equation by 5 and the second by 3: $$20x + 15y = 1100$$ $$6x + 15y = 750$$ Subtract the second from the first: $$14x = 350$$ $$x = 25$$ Substitute $$x = 25$$ into $$4x + 3y = 220$$: $$100 + 3y = 220$$ $$3y = 120$$ $$y = 40$$ Point: $$(25, 40)$$ 3. **Intersection of** $$3x + 4y = 270$$ **and** $$2x + 5y = 250$$: Multiply the first equation by 5 and the second by 4: $$15x + 20y = 1350$$ $$8x + 20y = 1000$$ Subtract the second from the first: $$7x = 350$$ $$x = 50$$ Substitute $$x = 50$$ into $$3x + 4y = 270$$: $$150 + 4y = 270$$ $$4y = 120$$ $$y = 30$$ Point: $$(50, 30)$$ ### Step 4: Evaluate the cost at each intersection point 1. $$(10, 60)$$: $$\text{Cost} = 15(10) + 12(60) = 150 + 720 = 870$$ 2. $$(25, 40)$$: $$\text{Cost} = 15(25) + 12(40) = 375 + 480 = 855$$ 3. $$(50, 30)$$: $$\text{Cost} = 15(50) + 12(30) = 750 + 360 = 1110$$ ### Step 5: Determine the minimum cost The minimum cost is obtained at the intersection point $$(25, 40)$$ with a cost of $$\$855$$. Therefore, the minimum cost to meet the picnic organizer's requirements is: $$\boxed{855 \text{ dollars}}$$ Would you like more details or have any questions? Here are 8 related questions to expand on this topic: 1. How can we verify the solution graphically? 2. What happens if the prices of the platters change? 3. How can we handle different types of constraints in linear programming? 4. What are some real-life applications of linear programming? 5. How does the Simplex Method solve such problems? 6. Can we solve this problem using integer programming for exact platter counts? 7. What if there are more than two types of platters available? 8. How can we optimize for other factors like quality or variety along with cost? **Tip:** Linear programming problems can often be visualized graphically for two variables, which helps in understanding the feasible region and optimal solutions.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Programming
Inequalities
Optimization
Formulas
Cost function: Cost = 15x + 12y
Theorems
-
Suitable Grade Level
Advanced High School
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