13.21
A function of two variables can reach maximum or minimum values at specific points on its surface.
A local maximum appears when the function value is higher than all nearby points, while a local minimum appears when it is lower.
These extremes happen at critical points, where the tangent plane is locally flat.
To find them, calculate the partial derivatives with respect to both variables. A point is critical when both derivatives are zero, indicating no change in any direction.
For example, imagine 12 square meters of cardboard is given to build the largest possible open-top box. While the volume depends on length x, width y, and height z, the limited material acts as a constraint.
This constraint allows height z to be expressed as a function of x and y, reducing the volume formula to just two variables.
The goal is to find the maximum volume, which occurs at a critical point where the volume function is locally flat.
To locate this point, the partial derivatives of V on x and y are calculated and set to zero. Solving these resulting equations gives the length and width. Substituting these values into the volume function gives the maximum volume.
In multivariable calculus, a function of two variables can exhibit local maximum or minimum values at certain points on its surface. A local maximum occurs when the function's value at a point is greater than at all nearby points, while a local minimum occurs when the function’s value is less than at all nearby locations. These points are referred to as local extrema and are of central importance in optimization problems.
Local extrema are found at critical points, where the surface becomes momentarily flat in all directions. At such points, the rate of change of the function with respect to each variable—called the partial derivatives—is zero. A point is considered critical if both the partial derivative with respect to the first variable and the partial derivative with respect to the second variable are equal to zero. This means that small movements in either direction do not cause the function's value to change.
To determine whether a critical point represents a maximum, minimum, or saddle point, additional methods such as examining the second-order partial derivatives or using the Hessian matrix are required.
As an application, consider maximizing the volume of an open-top rectangular box constructed from 12 square meters of cardboard. The box’s volume depends on three dimensions—length, width, and height—but the total surface area provides a constraint. By expressing volume in terms of two variables, such as length and width, and incorporating the constraint into the model, one can apply optimization techniques.
Partial derivatives of the resulting volume function with respect to each variable are calculated and set to zero to locate the critical points. Solving these conditions yields the box dimensions that provide the maximum volume. At the critical point, the function attains a local maximum, and the box reaches an optimal volume of 4 cubic meters. This process illustrates how the concept of critical points is used to solve practical optimization problems involving multiple variables and constraints.
A function of two variables can reach maximum or minimum values at specific points on its surface.
A local maximum appears when the function value is higher than all nearby points, while a local minimum appears when it is lower.
These extremes happen at critical points, where the tangent plane is locally flat.
To find them, calculate the partial derivatives with respect to both variables. A point is critical when both derivatives are zero, indicating no change in any direction.
For example, imagine 12 square meters of cardboard is given to build the largest possible open-top box. While the volume depends on length x, width y, and height z, the limited material acts as a constraint.
This constraint allows height z to be expressed as a function of x and y, reducing the volume formula to just two variables.
The goal is to find the maximum volume, which occurs at a critical point where the volume function is locally flat.
To locate this point, the partial derivatives of V on x and y are calculated and set to zero. Solving these resulting equations gives the length and width. Substituting these values into the volume function gives the maximum volume.
From Chapter 13:
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