Problem 8 - Active Calculus Multivariable

Question

Find \(A\) and \(B\) so that \(f(x, y)=x^{2}+A x+y^{2}+B\) has a local minimum at the point \((7,0),\) with \(z\) -coordinate 5 . \(A=\) ___________. \(B=\) ___________.

Step-by-Step Solution

Answer

The values of \(A\) and \(B\) that make the function \(f(x, y) = x^2 + Ax + y^2 + B\) have a local minimum at the point \((7, 0)\) with \(z\)-coordinate 5 are: \(A = -14\) \(B = 54\)

Step 1: Calculate the partial derivatives

First, we need to find the first-order partial derivatives of \(f(x, y)\) with respect to both \(x\) and \(y\). The partial derivatives are given by: \(f_x(x, y) = \frac{\partial f}{\partial x} = 2x + A\) \(f_y(x, y) = \frac{\partial f}{\partial y} = 2y\)

Step 2: Use the given point for local minimum

Since we are given a point \((7, 0)\) where there is a local minimum, we will substitute these values of \(x\) and \(y\) into the partial derivatives, and set them equal to zero. This is because at the local minimum, the first-order partial derivatives are equal to zero. \(f_x(7, 0) = 2(7) + A = 0\) \(f_y(7, 0) = 2(0) = 0\)

Step 3: Find the value of A

To find the value of \(A\), we can solve the equation obtained in step 2 for \(A\): \(2(7) + A = 0\) \(A = -2(7) = -14\)

Step 4: Use the given z-coordinate

We are given the \(z\)-coordinate of the local minimum as 5. Therefore, we can substitute the given values of \(x\) and \(y\) and the calculated value of \(A\) into the function, and set it equal to 5: \(f(7, 0) = (7)^2 -14(7) + (0)^2 + B = 5\)

Step 5: Find the value of B

To find the value of \(B\), we can solve the equation obtained in step 4 for \(B\): \(49 - 98 + B = 5\) \(B = 5 - 49 + 98 = 54\)

Step 6: Final Answer

The values of \(A\) and \(B\) that make the function \(f(x, y) = x^2 + Ax + y^2 + B\) have a local minimum at the point \((7, 0)\) with \(z\)-coordinate 5 are: \(A = -14\) \(B = 54\)

Key Concepts

Partial Derivatives

In the realm of multivariable calculus, a function may depend on two or more variables, say, x and y. To explore how this function behaves with respect to one variable while keeping the others constant, we calculate what's known as the partial derivative. These computations reveal the function's rate of change in a specific direction.

Let's consider a function z = f(x, y). The partial derivative with respect to x is denoted by f_x(x, y) or \(\frac{\partial f}{\partial x}\). Similarly, the derivative with respect to y is \(f_y(x, y)\) or \(\frac{\partial f}{\partial y}\). To find these derivatives, we treat y as a constant when differentiating with respect to x, and vice versa.

It's akin to examining the function's slope along the x-axis and y-axis independently. An understanding of partial derivatives is crucial for identifying the character of points on the function's surface, such as local maxima, minima, or saddle points. By calculating and studying partial derivatives, we gain insights into the function's local behavior and directional tendencies.

First-order Conditions

Closely linked to the concept of partial derivatives is that of first-order conditions, sometimes also referred to as the 'stationary conditions'. These conditions are employed for locating the critical points of a function where the function could potentially have a local maximum or minimum, or a saddle point.

According to the first-order conditions, at a critical point, all the first-order partial derivatives of the function must be equal to zero—essentially, the surface becomes flat in all directions at that point. The mathematical expression can be written as \(f_x(x, y) = 0\) and \(f_y(x, y) = 0\) for a function f(x, y).

This 'flattening' occurs because, at the extreme points (local minima/maxima), the function neither increases nor decreases as one moves infinitesimally in any direction, which is implied by its tangent plane being horizontal. Verifying this condition is usually the starting point in the process of finding local minimum and maximum values of multivariable functions.

Optimization in Multivariable Calculus

When it comes to optimization in multivariable calculus, the goal is to determine the points at which a function reaches its lowest or highest values under a set of constraints, if any. After identifying the first-order conditions, if additional analysis validates a local minimum, then the optimization problem is partially solved—subject to final confirmation by second-order conditions.

In multivariable calculus, optimization isn't just about finding a single value but often involves finding the most suitable set of values for variables that cause a function to reach its extreme value. In practical applications, optimization helps in making efficient decisions—be it minimizing costs or maximizing profits in business, or optimizing functions representing physical phenomena in engineering and natural sciences.

For instance, to optimize the function f(x, y) = x^2 + Ax + y^2 + B, we applied the first-order conditions to find the necessary constants A and B. This is a preliminary step in optimization, ensuring that a function behaves as desired at specific points, which in this case was to have a local minimum at a given point with a certain z-coordinate. It's a fascinating interplay between algebraic manipulation and the geometric understanding of surfaces and curves in higher dimensions.