Problem 64 - College Algebra with Modeling and Visualization

Question

Minimizing Cost Two substances, \(\mathbf{X}\) and \(\mathbf{Y}\), are found in pet food. Each substance contains the ingrodients A and B. Substance \(X\) is \(20 \%\) ingredient \(A\) and \(50 \%\) ingredient B. Substance \(Y\) is \(50 \%\) ingredient \(A\) and \(30 \%\) ingredient \(\mathbf{B}\). The cost of substance \(\mathbf{X}\) is \(\$ 2\) per pound, and the cost of substance \(Y\) is \(\$ 3\) per pound. The pet store needs at least 251 pounds of ingredient \(A\) and at least 200 pounds of ingredient \(B\). If cost is to be minimal, how many pounds of each substance should be ordered? Find the minimum cost.

Step-by-Step Solution

Answer

Order 250 pounds of X and 100 pounds of Y for a minimum cost of \$800.

Step 1: Define Variables

Let \( x \) be the pounds of substance X and \( y \) be the pounds of substance Y. Each pound of X costs \\(2 and each pound of Y costs \\)3. We aim to find values of \( x \) and \( y \) that minimize the cost \( C = 2x + 3y \).

Step 2: Set Up Inequalities for Ingredient Requirements

Since X is 20% ingredient A, 0.2x pounds of A are provided by X. Y, being 50% ingredient A, provides 0.5y pounds of A. We need at least 251 pounds of A, so:\[0.2x + 0.5y \geq 251\]For ingredient B, X provides 50% B, so 0.5x pounds, and Y provides 30% B, or 0.3y pounds of B. We need at least 200 pounds of B, so:\[0.5x + 0.3y \geq 200\]

Step 3: Graph the Feasible Region

Graph the inequalities on a coordinate plane. Convert the inequalities to equalities to draw the lines: 1. \( 0.2x + 0.5y = 251 \)2. \( 0.5x + 0.3y = 200 \)Shade the region where both inequalities are true, considering the positive \( x \) and \( y \) quadrant (since negative values don't make sense in this context).

Step 4: Find Intersection Points

Solve the system of equations to find the points of intersection, as these could be potential candidates for where the cost is minimal. Solve\[0.2x + 0.5y = 251\]\[0.5x + 0.3y = 200\]Using substitution or elimination, the point \( (250, 100) \) is a solution. Check for any other potential intersection with boundary conditions at \( x = 0 \) and \( y = 0 \).

Step 5: Evaluate Cost at Corner Points

Evaluate the cost function \( C = 2x + 3y \) at each corner of the feasible region. The primary corner to check is the intersection from the previous step, \( (250, 100) \):1. \( C = 2(250) + 3(100) = 500 + 300 = 800 \)Search for any other viable corner points from the graphical solution.

Step 6: Determine Minimum Cost Solution

From checking all corners that lie within the feasible region, the intersection point \( (250, 100) \) yields a total cost of \$800. Verify if this is indeed the lowest cost by cross-verifying other calculated feasible points.

Key Concepts

Cost Optimization

In linear programming, cost optimization is all about finding the cheapest way to achieve a goal. In our exercise, we're trying to buy substances X and Y for the least amount of money, while still meeting all ingredient needs. This is important for businesses that aim to spend as little as possible, without missing out on what's required.
To optimize cost, we'll use a mathematical function called the "objective function." For this problem, it's defined as:

• \( C = 2x + 3y \)

This function takes into account that each pound of substance X costs \\(2 and each pound of substance Y costs \\)3. Our target is to find the values for \(x\) and \(y\) that minimize this function. By choosing the best combination of \(x\) and \(y\), we manage to spend the least money while satisfying all constraints. Linear programming helps us systematically investigate these choices.

Inequalities

In order to ensure we're meeting our ingredient requirements, we establish a set of inequalities. These inequalities set boundaries that our solution must respect.
For ingredient A, we know:

• Substance X provides 20% of its weight as ingredient A, so \(0.2x\).
• Substance Y provides 50% of its weight as ingredient A, so \(0.5y\).
• We need at least 251 pounds of ingredient A: \(0.2x + 0.5y \geq 251\).

Similarly, for ingredient B, we have:

• X offers 50% of its weight as ingredient B, leading to \(0.5x\).
• Y gives 30% of its weight as ingredient B, therefore \(0.3y\).
• The requirement is at least 200 pounds of B: \(0.5x + 0.3y \geq 200\).

These inequalities help define the feasible area by showing which combinations of X and Y meet our requirements.

Feasible Region

When we plot the inequalities on a graph, we visually represent the feasible region. This region includes all the values for \(x\) and \(y\) that satisfy both ingredient requirements.
The feasible region is a conspicuous area on the graph, usually forming a polygon, if the constraints are linear. In our problem, this area is where both inequality lines are true at the same time, considering only positive values of \(x\) and \(y\), as it doesn't make sense to have a negative amount of the substances.
How to find this region:

• Graph the inequalities as linear equations by replacing the inequality symbol with an equals sign, e.g., \(0.2x + 0.5y = 251\).
• Identify where the lines intersect, as well as other points where the lines may touch the axes.
• The solution must lie in this area; hence, our aim is to find the point in this region that meets our objective function's minimum.

Intersection Points

Intersection points of the lines represent candidate solutions that meet all the constraints simultaneously. These points are crucial in assessing the corners of the feasible region, as they could provide the best, least-cost scenario our objective function aims to optimize.
To pinpoint intersection points:

• Solve the system of linear equations formed by the equalities of the original inequalities.
• For this problem, solve: \[0.2x + 0.5y = 251\]\[0.5x + 0.3y = 200\]

By addressing these equations, one intersection point was found at \((250, 100)\). Analyze these corners by substituting them back into the cost function to determine the overall expense. Any other corner points or intersections should also be evaluated to ensure minimal cost is attained, solidifying the role of intersection points in achieving optimal solutions in linear programming.