91Ó°ÊÓ

In multivariable calculus, critical points are essential in the study of functions involving more than one variable. For a function of two variables, like the profit model given, critical points occur where the partial derivatives with respect to each variable are zero. These are points where the function could have a local maximum, a local minimum, or a saddle point.

To find the critical points of a function like \( P(x, y) = -2x^2 + 40x + 100y - 5y^2 \), we:

• Calculate the partial derivatives
• Set these derivatives to zero and solve the resulting equations

For our example, the partial derivative with respect to \( x \) is \( \frac{\partial P}{\partial x} = -4x + 40 \), and for \( y \), it is \( \frac{\partial P}{\partial y} = 100 - 10y \). Setting \( \frac{\partial P}{\partial x} \) and \( \frac{\partial P}{\partial y} \) equal to zero provides functions in \( x \) and \( y \) which, when solved, give the critical point \( (x, y) = (10, 10) \).

Finding the critical points is the first step in determining the behavior of a multivariable function.

The second derivative test in multivariable calculus helps determine whether a critical point is a local maximum, local minimum, or saddle point. For a function of two variables, this involves examining the second partial derivatives and calculating the Hessian determinant.

For the function \( P(x, y) = -2x^2 + 40x + 100y - 5y^2 \), the second partial derivatives are:

• \( f_{xx} = \frac{\partial^2 P}{\partial x^2} = -4 \)
• \( f_{yy} = \frac{\partial^2 P}{\partial y^2} = -10 \)
• \( f_{xy} = \frac{\partial^2 P}{\partial x \partial y} = 0 \)

The Hessian matrix \( H \) is then \[H = \begin{vmatrix}f_{xx} & f_{xy} \f_{xy} & f_{yy}\end{vmatrix} = \begin{vmatrix}-4 & 0 \0 & -10\end{vmatrix} = 40\]
The test states that if the determinant of the Hessian is positive and both \( f_{xx} \) and \( f_{yy} \) are negative, the critical point is a local maximum. In this case, since \( f_{xx} < 0 \), \( f_{yy} < 0 \), and \( H > 0 \), the critical point at \( (10, 10) \) is indeed a point of local maximum.

Partial derivatives are fundamental in understanding functions involving multiple variables, like in our profit function \( P(x, y) = -2x^2 + 40x + 100y - 5y^2 \). They represent the rate at which the function changes with respect to one variable, while keeping the other constant.

To compute the partial derivatives for this function:

• With respect to \( x \): The partial derivative is \( \frac{\partial P}{\partial x} = -4x + 40 \). This represents the change in profit when increasing the hours logged by desktop engineers, while holding network engineers' hours constant.
• With respect to \( y \): The partial derivative is \( \frac{\partial P}{\partial y} = 100 - 10y \). This indicates the change in profit when increasing the hours logged by network systems engineers, keeping desktop engineers' hours constant.

By setting these partial derivatives to zero, we identify the critical points. This helps in understanding how changes in one variable affect the function, allowing us to optimize the output—in this case, maximizing the profit for the firm.