Problem 5 - A First Course in Mathematical Modeling

Question

A company manufactures \(x\) floor lamps and \(y\) table lamps each day. The profit in dollars for the manufacture and sale of these lamps is $$ P(x, y)=18 x+2 y-0.05 x^{2}-0.03 y^{2}+0.02 x y-100 $$ Find the daily production level of each lamp to maximize the company's profits.

Step-by-Step Solution

Answer

Produce approximately 183 floor lamps and 42 table lamps daily.

Step 1: Find the First Derivatives

To find the production levels that maximize profit, we start by taking the partial derivatives of the profit function \(P(x, y)\) with respect to \(x\) and \(y\). Calculating these, we find:\[P_x = \frac{\partial P}{\partial x} = 18 - 0.1x + 0.02y\]\[P_y = \frac{\partial P}{\partial y} = 2 - 0.06y + 0.02x\]

Step 2: Set Derivatives to Zero

We set both partial derivatives equal to zero in order to find the critical points where the profit could be maximized.\[ 18 - 0.1x + 0.02y = 0 \]\[ 2 - 0.06y + 0.02x = 0 \]

Step 3: Solve the System of Equations

Solve the system of equations from Step 2. Start by solving one equation for one variable and substitute into the other.1. Solve \(18 = 0.1x - 0.02y\) \[ x = \frac{18 + 0.02y}{0.1} \]2. Substitute \(x\) into the second equation: \[ 2 = 0.06y - 0.02\left(\frac{18 + 0.02y}{0.1}\right) \] Simplifying leads to: \[ 2 = 0.06y - 0.36 - 0.004y \] \[ 2.36 = 0.056y \] Solve for \(y\): \[ y = \frac{2.36}{0.056} \approx 42.14 \]

Step 4: Substitute Back to Find x

Use the value of \(y\) to find \(x\). Substitute \(y = 42.14\) into the expression for \(x\):\[ x = \frac{18 + 0.02(42.14)}{0.1} \]Calculate to find \(x\):\[ x \approx 182.8 \]

Step 5: Check Second Derivatives (Second Derivative Test)

Calculate the second partial derivatives to use the second derivative test:\[ P_{xx} = -0.1, \quad P_{yy} = -0.06, \quad P_{xy} = 0.02 \]The determinant of the Hessian matrix is:\[ H = P_{xx}P_{yy} - (P_{xy})^2 = (-0.1)(-0.06) - (0.02)^2 = 0.006 - 0.0004 = 0.0056 \] Since \(H > 0\) and \(P_{xx} < 0\), the critical point is a maximum. Thus, the maximum profit occurs when \(x \approx 182.8\) and \(y \approx 42.14\), which means rounding to whole numbers: approximately 183 floor lamps and 42 table lamps.

Key Concepts

Partial Derivatives

Partial derivatives are fundamental tools in optimization, particularly in economics when dealing with functions that depend on multiple variables. A partial derivative measures how a function changes as one of its variables changes, while keeping the other variables constant.
In the context of our lamp example, the profit function \( P(x, y) \) depends on two variables: the number of floor lamps \( x \) and the number of table lamps \( y \). To determine how the profit changes as we vary \( x \) and \( y \) independently, we calculate two partial derivatives:

• \( P_x = \frac{\partial P}{\partial x} \): Measures the rate of change of profit with respect to the number of floor lamps, keeping the number of table lamps constant.
• \( P_y = \frac{\partial P}{\partial y} \): Measures the rate of change of profit with respect to the number of table lamps, keeping the number of floor lamps constant.

These derivatives help us identify how sensitive overall profit is to changes in either type of lamp produced.

Critical Points

When optimizing a function like the profit function, critical points are points where the first derivatives (partial derivatives) are zero. This is where the function could potentially have a maximum, a minimum, or a saddle point.
For the lamp production problem, we set the partial derivatives \( P_x \) and \( P_y \) to zero to locate the critical points:

• \( 18 - 0.1x + 0.02y = 0 \)
• \( 2 - 0.06y + 0.02x = 0 \)

Solving these equations gives us the values of \( x \) and \( y \) at the critical point. These values are not guaranteed to be optimum, so further analysis, such as the second derivative test, is needed to confirm their nature.

Second Derivative Test

The second derivative test is a method to determine the nature of critical points found by solving the first derivative conditions. It involves calculating the second partial derivatives and evaluating the Hessian matrix.
In the lamp problem, after identifying the critical points, we calculate:

• \( P_{xx} = \frac{\partial^2 P}{\partial x^2} = -0.1 \)
• \( P_{yy} = \frac{\partial^2 P}{\partial y^2} = -0.06 \)
• \( P_{xy} = \frac{\partial^2 P}{\partial x \partial y} = 0.02 \)

The results are used to form the Hessian matrix, and its determinant helps to determine if the critical point is a maximum, minimum, or saddle point.

Hessian Matrix

The Hessian matrix is a square matrix comprising the second partial derivatives of a function with respect to its variables. It's a key tool in multivariable calculus for analyzing critical points.
The Hessian matrix \( H \) for the given profit function is:\[H = \begin{bmatrix}P_{xx} & P_{xy} \P_{xy} & P_{yy} \end{bmatrix} = \begin{bmatrix}-0.1 & 0.02 \0.02 & -0.06 \end{bmatrix}\]To apply the second derivative test, we compute the determinant of this matrix:\[H = P_{xx}P_{yy} - (P_{xy})^2\]For the lamp example, the determinant is \( 0.0056 \), which is positive. If \( H > 0 \) and \( P_{xx} < 0 \), the critical point corresponds to a local maximum. This allows us to conclude that the production of about 183 floor lamps and 42 table lamps will maximize profit.