Problem 16 - Calculus and its applications

Question

Assume that relative maximum and minimum values are absolute maximum and minimum values. A concert promoter produces two kinds of souvenir shirt. Total revenue from the sale of \(x\) thousand shirts at \(\$ 18\) each and \(y\) thousand at \(\$ 25\) each is given by $$ R(x, y)=18 x+25 y $$ The company determines that the total cost, in thousands of dollars, of producing \(x\) thousand of the \(\$ 18\) shirt and \(y\) thousand of the \(\$ 25\) shirt is $$ C(x, y)=4 x^{2}-6 x y+3 y^{2}+20 x+19 y-12 $$ How many of each type of shirt must be produced and sold to maximize profit?

Step-by-Step Solution

Answer

Produce and sell 2,000 of the $18 shirts and 3,000 of the $25 shirts to maximize profit.

Step 1: Define the Profit Function

The profit function is defined as the difference between the total revenue and total cost. Thus, we write the profit function as follows: \[ P(x, y) = R(x, y) - C(x, y) = (18x + 25y) - (4x^2 - 6xy + 3y^2 + 20x + 19y - 12) \]Simplify the equation: \[ P(x, y) = -4x^2 + 6xy - 3y^2 - 2x + 6y + 12 \]

Step 2: Find the Critical Points

To find the critical points of the profit function, we need to take the partial derivatives \( P_x(x, y) \) and \( P_y(x, y) \) and set them equal to zero.First, find \( P_x(x, y) \):\[ P_x(x, y) = \frac{\partial}{\partial x}(-4x^2 + 6xy - 3y^2 - 2x + 6y + 12) = -8x + 6y - 2 \]Set \( P_x(x, y) = 0 \):\[ -8x + 6y - 2 = 0 \quad \text{(Equation 1)} \]Next, find \( P_y(x, y) \):\[ P_y(x, y) = \frac{\partial}{\partial y}(-4x^2 + 6xy - 3y^2 - 2x + 6y + 12) = 6x - 6y + 6 \]Set \( P_y(x, y) = 0 \):\[ 6x - 6y + 6 = 0 \quad \text{(Equation 2)} \]

Step 3: Solve the System of Equations

Now, solve the system of equations obtained from the partial derivatives:1. \( -8x + 6y - 2 = 0 \)2. \( 6x - 6y + 6 = 0 \)From Equation 2: \[ 6x - 6y + 6 = 0 \rightarrow 6x = 6y - 6 \rightarrow x = y - 1 \]Substitute \( x = y - 1 \) into Equation 1:\[ -8(y - 1) + 6y - 2 = 0 \rightarrow -8y + 8 + 6y - 2 = 0 \rightarrow -2y + 6 = 0 \rightarrow 2y = 6 \rightarrow y = 3 \]Substitute \( y = 3 \) back into \( x = y - 1 \):\[ x = 3 - 1 = 2 \]

Step 4: Verify Second Derivative Test

To ensure this point is a maximum, use the second derivative test, which involves the Hessian matrix. Calculate the second derivatives:\[ P_{xx} = -8, \quad P_{yy} = -6, \quad P_{xy} = 6 \]The Hessian determinant is given by:\[ H = \begin{vmatrix} P_{xx} & P_{xy} \ P_{xy} & P_{yy} \end{vmatrix} = (-8)(-6) - (6)(6) = 48 - 36 = 12 \]Since \( H > 0 \) and \( P_{xx} < 0 \), the critical point \((2, 3)\) is a local maximum.

Key Concepts

Partial Derivatives

Partial derivatives are crucial when dealing with functions of multiple variables, like in our profit maximization problem with variables \(x\) and \(y\). When we have a function \(P(x, y)\), which represents profit in this case, partial derivatives let us see how changes in one variable affect the profit while keeping the other variable constant.

To find the partial derivative of \(P\) with respect to \(x\), denoted by \(P_x(x, y)\), we treat \(y\) as a constant and differentiate \(P\) with respect to \(x\). Similarly, for \(P_y(x, y)\), \(x\) is treated as a constant while differentiating with respect to \(y\).

• \(P_x(x, y) = \frac{\partial}{\partial x}(-4x^2 + 6xy - 3y^2 - 2x + 6y + 12) = -8x + 6y - 2\)
• \(P_y(x, y) = \frac{\partial}{\partial y}(-4x^2 + 6xy - 3y^2 - 2x + 6y + 12) = 6x - 6y + 6\)

By setting these derivatives equal to zero, we find critical points which are potential candidates for maximum or minimum values of the function.

Critical Points

Critical points are where the partial derivatives of the function are zero. These points are key in determining where a function might reach a local maximum or minimum. In context, for the profit function \(P(x, y)\), we find critical points by solving the following system of equations:

• \(-8x + 6y - 2 = 0\)
• \(6x - 6y + 6 = 0\)

These equations result from setting the partial derivatives equal to zero. Solving them allows us to find specific values for \(x\) and \(y\) where the function's slope is flat in all directions.

For instance, solving these gives us \(x = 2\) and \(y = 3\). Critical points don’t automatically mean we've found a maximum; further analysis is needed. This is where the second derivative test and Hessian matrix come into play, providing a deeper insight into the nature of these points.

Second Derivative Test

Once we have the critical points, the second derivative test helps to determine their nature—whether they are maxima, minima, or saddle points. For a function of two variables, the test utilizes the Hessian matrix.

The second derivatives required for this are:

• \(P_{xx} = \frac{\partial^2}{\partial x^2} P(x, y) = -8\)
• \(P_{yy} = \frac{\partial^2}{\partial y^2} P(x, y) = -6\)
• \(P_{xy} = \frac{\partial^2}{\partial x \partial y} P(x, y) = 6\)

With these, we can compute the Hessian determinant to apply the test. If the determinant is positive and \(P_{xx}\) is less than zero, as in this exercise, the point is a local maximum. In our example, the determinant value confirms that the critical point \((2, 3)\) is indeed a local maximum where the promoter's profit is maximized.

Hessian Matrix

The Hessian matrix is an essential construction in multivariable calculus for analyzing the curvature near critical points. It is a square matrix composed of the second-order partial derivatives of a function.

For the profit function \(P(x, y)\), the Hessian is structured as:

• \[H = \begin{pmatrix} P_{xx} & P_{xy} \ P_{xy} & P_{yy} \end{pmatrix}\]
• Here, \(P_{xx} = -8\), \(P_{yy} = -6\), and \(P_{xy} = 6\).

The Hessian determinant is calculated as:
\[H = (-8)(-6) - (6)(6) = 48 - 36 = 12\]
The positive Hessian determinant and negative \(P_{xx}\) indicate that the critical point \((2, 3)\) is a local maximum.

This tool allows us to clearly interpret the behavior of the function around critical points, simplifying the analysis of complex systems and assisting in the decision-making process.