Problem 35 - Calculus and its applications

Question

A manufacturer of decorative end tables produces two models, basic and large. Its weekly profit function is modeled by $$ P(x, y)=-x^{2}-2 y^{2}-x y+140 x+210 y-4300 $$ where \(x\) is the number of basic models sold each week and \(y\) is the number of large models sold each week. The warehouse can hold at most 90 tables. Assume that \(x\) and \(y\) must be nonnegative. How many of each model should be produced to maximize weekly profit, and what will the maximum profit be?

Step-by-Step Solution

Answer

Produce 60 basic models and 30 large models for a maximum weekly profit of $1600.

Step 1: Understand the problem

We need to find the combination of basic models \(x\) and large models \(y\) that maximizes the profit function \(P(x, y)\). We observe the constraints: maximum 90 tables total, and \(x\) and \(y\) are nonnegative.

Step 2: Express the constraints

The constraint for the warehouse is \(x + y \leq 90\). Additionally, \(x\geq0\) and \(y\geq0\). These inequalities encapsulate the feasible region for our optimization problem.

Step 3: Calculate partial derivatives

To find the critical points that could maximize the profit, we first calculate the partial derivatives of the function: \(\frac{\partial P}{\partial x} = -2x - y + 140\) and \(\frac{\partial P}{\partial y} = -4y - x + 210\). Set these to zero to find critical points.

Step 4: Solve for critical points

Solving the system of equations from the partial derivatives: \(-2x - y + 140 = 0\) and \(-4y - x + 210 = 0\). Solving these simultaneously provides \(x = 70\) and \(y = 35\).

Step 5: Verify if the critical point Feasible

Check the constraints with \(x=70\) and \(y=35\): \(x+y = 105\) does not satisfy \(x+y\leq90\). Hence, the point \((70, 35)\) is not feasible.

Step 6: Evaluate at the vertices of the feasible region

We test the feasible boundary conditions to find potential maximum points. Check \((x, y) = (90,0), (0,90), (60,30)\). Calculate \(P(x,y)\) for each:

Step 7: Calculate optimal results

Calculate the profit at the feasible points: - For \((90,0)\), \(P(90,0) = -8100 + 12600 - 4300 = 1200\) - For \((0,90)\), \(P(0,90) = -16200 + 18900 - 4300 = -1600\) - For \((60,30)\), \(P(60,30) = -7200 - 1800 - 1800 + 8400 + 6300 - 4300 = 1600\).

Step 8: Select the maximum profit

Among the calculated profits, the maximum is \(1600\) at \((60, 30)\). Thus, producing 60 basic models and 30 large models yields the highest profit.

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept when working with functions that have more than one variable—like the profit function in our problem, which depends on both the number of basic and large table models. In this context, the partial derivative of a function with respect to one variable holds the other variable constant. This allows us to assess how changes in one variable impact the output.

To identify where a function's output might be maximized (or minimized), we look to where these partial derivatives equal zero. These points, known as critical points, give us a foundation to explore potential maxima or minima of the function. For our profit function, the partial derivatives are:

• The partial derivative with respect to x: \(\frac{\partial P}{\partial x} = -2x - y + 140\).
• The partial derivative with respect to y: \(\frac{\partial P}{\partial y} = -4y - x + 210\).

Finding where these two partial derivatives are equal to zero allows us to explore the critical points essential for maximizing profit.

Optimization Constraints

Constraints are conditions that must be satisfied for any proposed solution to be considered valid. In optimization problems like ours, constraints define the limits within which the solution must lie. Here, our constraints come from the physical limitation of the warehouse and the non-negative requirement on tables:

• The warehouse constraint: \(x + y \leq 90\) ensures you don't exceed capacity.
• The non-negativity constraints: \(x \geq 0\) and \(y \geq 0\) ensure we can't have negative quantities of tables produced.

These constraints outline what is known as the feasible region—an area within which any solution must fall.

All calculations, including finding potential maxima, must respect these constraints to ensure a valid and implementable solution.

Critical Points

In optimization, critical points are specific points within the domain of the function where the rate of change of the function is zero. These points provide valuable insights into where potential maxima or minima might occur.

For our profit maximization problem, we find critical points by setting the partial derivatives equal to zero:

• \(-2x - y + 140 = 0\)
• \(-4y - x + 210 = 0\)

Solving this system of equations gives us \(x = 70\) and \(y = 35\). However, once found, it's crucial to check these points against the constraints to ensure they lie within our feasible region. For instance, even though \(x = 70\) and \(y = 35\) are mathematically derived as critical, they exceed the allowable warehouse capacity, proving them non-feasible in this problem.

Feasible Region

The feasible region in an optimization problem is the collection of all possible points that satisfy the problem's constraints. It's a critical concept since any optimal solution must lie within this specified area.

In our case, the feasible region is defined by the following inequalities:

• \(x + y \leq 90\)
• \(x \geq 0\)
• \(y \geq 0\)

This region is often represented as an area on a graph where solutions can "live," bounded by these constraints. For this problem, several key points on the edges of this region are tested. They are boundary cases like \(x = 90, y = 0\) or \(x = 60, y = 30\), which are practical to test for maximization.

The feasible region thus acts like a filter, narrowing down viable solution candidates by eliminating those that don't comply with all constraints.