Problem 355 - Calculus Volume 3

Question

A company manufactures two types of athletic shoes: jogging shoes and cross-trainers. The total revenue from \(x\) units of jogging shoes and \(y\) units of cross-trainers is given by \(R(x, y)=-5 x^{2}-8 y^{2}-2 x y+42 x+102 y\), where \(x\) and \(y\) are in thousands of units. Find the values of \(x\) and \(y\) to maximize the total revenue.

Step-by-Step Solution

Answer

The values of \( x \) and \( y \) to maximize the revenue are \( x = 3 \) and \( y = 6 \).

Step 1: Identify the Function Type

The given function \( R(x, y) = -5x^{2} - 8y^{2} - 2xy + 42x + 102y \) is a quadratic function in two variables, a type of multivariable function where the goal is to find the extreme (either maximum or minimum) points.

Step 2: Determine Derivatives

Calculate the first partial derivatives of \( R(x, y) \) with respect to \( x \) and \( y \). For \( x \):\[ \frac{\partial R}{\partial x} = -10x - 2y + 42 \]For \( y \):\[ \frac{\partial R}{\partial y} = -16y - 2x + 102 \]

Step 3: Find Critical Points

Set the partial derivatives to zero and solve the system of equations:1. \( -10x - 2y + 42 = 0 \)2. \( -16y - 2x + 102 = 0 \)Solving these simultaneously gives the critical points.

Step 4: Solve the System of Equations

From equation 1: \[ -10x - 2y + 42 = 0 \] Reformulate to find \( y \): \[ 2y = 42 - 10x \]\[ y = 21 - 5x \]Substitute \( y = 21 - 5x \) into equation 2:\[ -16(21-5x) - 2x + 102 = 0 \] \[ -336 + 80x - 2x + 102 = 0 \] \[ 78x - 234 = 0 \]Solve for \( x \):\[ 78x = 234 \]\[ x = 3 \]Use \( x = 3 \) to solve for \( y \):\[ y = 21 - 5(3) = 6 \]

Step 5: Analyze the Critical Point

The critical point found is \( (3, 6) \). To confirm this as a maximum, we can analyze the second derivatives or utilize the discriminant of the quadratic function to check if it represents a maximum.

Step 6: Confirm Maximization

Confirming using the second partial derivatives:1. \( \frac{\partial^{2} R}{\partial x^{2}} = -10 \)2. \( \frac{\partial^{2} R}{\partial y^{2}} = -16 \)3. \( \frac{\partial^{2} R}{\partial x \partial y} = -2 \)Compute the Hessian determinant:\[ D = (-10)(-16) - (-2)^2 = 160 - 4 = 156 \]Since \( D > 0 \) and \( \frac{\partial^{2} R}{\partial x^{2}} < 0 \), the critical point is a local maximum.

Key Concepts

Partial Derivatives

Partial derivatives are tools used to understand how a function changes as its input variables vary. In multivariable calculus, they allow us to examine how a function behaves concerning one variable, holding the others constant. This has a crucial application in optimizing functions, like in our exercise where we're aiming to maximize revenue.

For a function of two variables, such as the total revenue function \( R(x, y) = -5x^{2} - 8y^{2} - 2xy + 42x + 102y \), calculating the partial derivatives involves differentiating with respect to one variable at a time.

• With respect to \( x \): \( \frac{\partial R}{\partial x} = -10x - 2y + 42 \)
• With respect to \( y \): \( \frac{\partial R}{\partial y} = -16y - 2x + 102 \)

These derivatives provide the rate of change of revenue with respect to each type of shoe produced.

By setting these derivatives to zero, you create equations that help locate critical points, which are potential points of maximum or minimum revenue.

Critical Points

Critical points are specific values of the variables where the function's rate of change is zero, indicating potential maxima or minima.

In order to find these points, you set the partial derivatives equal to zero:

• \( -10x - 2y + 42 = 0 \)
• \( -16y - 2x + 102 = 0 \)

Solving this system of equations helps locate the critical points of the revenue function.

Upon solving the equations, we find a critical point at \( (3, 6) \). This means that manufacturing 3 units of jogging shoes and 6 units of cross-trainers could potentially maximize the revenue.

However, identifying critical points alone doesn't tell us whether they are indeed maxima or minima; further analysis is required.

Hessian Determinant

The Hessian determinant is a valuable tool in determining the nature of a critical point found from a multivariable function. It helps us decide if the critical point at hand is a maximum, a minimum, or a saddle point.

The Hessian determinant uses second derivatives, as shown for the revenue function:

• \( \frac{\partial^{2} R}{\partial x^{2}} = -10 \)
• \( \frac{\partial^{2} R}{\partial y^{2}} = -16 \)
• \( \frac{\partial^{2} R}{\partial x \partial y} = -2 \)

With these, the Hessian determinant \( D \) is calculated:
\[ D = (-10)(-16) - (-2)^2 = 160 - 4 = 156 \]
Since \( D > 0 \) and \( \frac{\partial^{2} R}{\partial x^{2}} < 0 \), the critical point \( (3, 6) \) is a local maximum.

This confirms that producing 3 units of jogging shoes and 6 units of cross-trainers is indeed optimal for maximizing revenue.

Quadratic Functions

Quadratic functions in multivariable calculus, such as the revenue function \( R(x, y) = -5x^{2} - 8y^{2} - 2xy + 42x + 102y \), represent surfaces that can have peaks (maxima) or troughs (minima). These are straightforward examples of polynomial expressions of degree two.

Such functions typically take the form \( ax^2 + by^2 + cxy + dx + ey + f \), where each term indicates contributions from individual variables or their interaction. The coefficients determine the curvature and direction of the surface.

Quadratic functions are integral to optimization problems in calculus, providing a framework for analyzing how changes in input variables affect the outcome. They are crucial in fields like economics, physics, and engineering, where such relationships are often modeled.

In our exercise, we're tasked with optimizing a revenue function, showcasing how quadratic functions can be employed to derive meaningful and practical solutions in business contexts.