91Ó°ÊÓ

Partial derivatives are a fundamental tool in calculus, especially when dealing with functions of multiple variables, like our profit function, \( P(x, y) = -0.002x^2 + 20x + 12.8y - 0.05y^2 \). Instead of taking a derivative in one dimension, partial derivatives allow us to see how the function changes along each variable individually.

This means calculating the derivative of the function with respect to one variable while treating the other as a constant. In our problem:

• We compute \( \frac{\partial P}{\partial x} = -0.004x + 20 \)
• We compute \( \frac{\partial P}{\partial y} = 12.8 - 0.1y \)

Finding these derivatives is the first step in determining where profits can be maximized, by solving \( \frac{\partial P}{\partial x} = 0 \) and \( \frac{\partial P}{\partial y} = 0 \) to find critical points.

The second derivative test provides an effective way to determine the nature of critical points found by setting the first derivatives to zero. By looking into the second derivatives, we can assess whether a point is a minimum, maximum, or a saddle point.

For functions of multiple variables, the test involves the second partial derivatives:

• \( \frac{\partial^2 P}{\partial x^2} = -0.004 \)
• \( \frac{\partial^2 P}{\partial y^2} = -0.1 \)
• \( \frac{\partial^2 P}{\partial x \partial y} = 0 \)

These values help construct the Hessian matrix, which is crucial for further testing the nature of critical points.

The Hessian matrix consolidates the second partial derivatives into a coherent structure to help determine the concavity at critical points. Consider it a matrix of these derivatives:\[H = \begin{bmatrix} \frac{\partial^2 P}{\partial x^2} & \frac{\partial^2 P}{\partial x \partial y} \\frac{\partial^2 P}{\partial y \partial x} & \frac{\partial^2 P}{\partial y^2} \end{bmatrix}\]For our function, this translates to:\[H = \begin{bmatrix}-0.004 & 0 \0 & -0.1 \end{bmatrix}\]By calculating the determinant of this matrix, \( D = (-0.004)(-0.1) - (0)(0) = 0.0004 \), and noting that it is positive, we confirm a local maximum at the critical point where \( \frac{\partial^2 P}{\partial x^2} < 0 \).

This information is pivotal for verifying the maximizing condition for our profit function.

Critical points are where the first derivatives (partial derivatives, in the case of functions of multiple variables) equal zero. They are essential because they indicate where a function may achieve a local maximum, minimum, or change direction suddenly (saddle point).

For the profit function \( P(x, y) \), we found critical points by setting the partial derivatives to zero:

• \( -0.004x + 20 = 0 \) resolved to \( x = 5000 \)
• \( 12.8 - 0.1y = 0 \) resolved to \( y = 128 \)

This critical point, \( (5000, 128) \), identifies the combination of sucker and peppermint stick sales (in thousands of pounds) that potentially maximizes profit. Using the Hessian matrix and second derivative test, we verify this point is indeed a maximum, ensuring decisions made on selling strategy are sound.