Problem 24 - Advanced Engineering Mathematics

Question

In Problems, find a vector that gives the direction in which the given function increases most rapidly at the indicated point. Find the maximum rate. $$ f(x, y)=x y e^{x-y} ;(5,5) $$

Step-by-Step Solution

Answer

The direction of increase is (30, -20), and the maximum rate is \(10\sqrt{13}\).

Step 1: Find the Gradient of the Function

The gradient of the function \(f(x, y) = x y e^{x-y}\) is a vector that consists of all the first order partial derivatives of \(f\). Calculate \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\).For \(\frac{\partial f}{\partial x}\), using the product and chain rules, we have:\[\frac{\partial}{\partial x}(x y e^{x-y}) = y e^{x-y} + x y e^{x-y}\]Simplifying, \(\frac{\partial f}{\partial x} = y e^{x-y} + x y e^{x-y}\).For \(\frac{\partial f}{\partial y}\), we have:\[\frac{\partial}{\partial y}(x y e^{x-y}) = x e^{x-y} - x y e^{x-y}\]Simplifying, \(\frac{\partial f}{\partial y} = x e^{x-y} - x y e^{x-y}\).

Step 2: Evaluate the Gradient at the Point

Substitute \((x, y) = (5, 5)\) into the gradient vector \(abla f\).Calculate:\[\left(5 \, e^{5-5} + 5 \times 5 \times e^{5-5}, 5 \, e^{5-5} - 5 \times 5 \times e^{5-5}\right)\]Simplifying, the gradient at \((5, 5)\) is: \[\left(5 \times 1 + 25 \times 1, 5 \times 1 - 25 \times 1\right) = (30, -20)\].

Step 3: Determine the Direction of Maximum Increase

The direction in which the function \(f(x, y)\) increases most rapidly is given by the gradient vector. For the point \((5,5)\), the direction vector is \((30, -20)\). This vector indicates the direction of steepest ascent for the function.

Step 4: Calculate the Maximum Rate of Increase

Calculate the magnitude of the gradient vector to find the maximum rate of increase:\[\| abla f \| = \sqrt{(30)^2 + (-20)^2}\]\[= \sqrt{900 + 400}\]\[= \sqrt{1300}\]\[= 10\sqrt{13}\]Therefore, the maximum rate of increase of the function at the given point is \(10\sqrt{13}\).

Key Concepts

Partial Derivatives

Partial derivatives help us understand how a function of multiple variables changes as one particular variable changes, while others are kept constant. For the function \( f(x, y) = x y e^{x-y} \), calculating partial derivatives involves using calculus to find distinct derivatives with respect to \( x \) and \( y \).
Here's how it's done:

• We start with \( \frac{\partial f}{\partial x} \), using product and chain rules. The derivative tells us how \( f \) changes as \( x \) changes slightly, while \( y \) stays fixed.
• The process is repeated for \( \frac{\partial f}{\partial y} \), to find how \( f \) changes with respect to small changes in \( y \).

This helps create a gradient vector, which serves as a foundation to determine the function's behavior in different directions. Partial derivatives are essential for analyzing complex functions across multiple dimensions.

Rate of Increase

The rate of increase is crucial in determining how fast the function value changes in a particular direction. At a given point, the direction and magnitude of this increase provide insights into the function's behavior.
The steps involved include:

• Calculating the gradient vector at the point of interest, which gives the direction of the steepest increase.
• Finding the magnitude of the gradient vector, representing the maximum rate at which the function can increase.

In our example, applying these steps suggests that the maximum rate at which the function\( f(x, y) = x y e^{x-y} \) climbs at the point \((5,5)\) is \(10\sqrt{13}\). Understanding the rate of increase allows for better predictions of the function's behavior around specific points.

Steepest Ascent

Steepest ascent is identified by the direction where the function increases most rapidly. This direction is given by the gradient vector. The gradient points towards the most significant uphill path a function can take.
For the function \( f(x, y) = x y e^{x-y} \), calculated at point \((5,5)\), the gradient vector \((30, -20)\) shows the steepest ascent. This vector is essential for optimizing functions, guiding where adjustments can be most effective.
Following the path of steepest ascent is similar to climbing the steepest possible path up a hill, ensuring the fastest increase in elevation.

Function Optimization

Function optimization involves finding the maximum or minimum values of a function. This process often requires identifying places where the function exhibits maximum increase or decrease.
In the case of the function \( f(x, y) = x y e^{x-y} \), by using concepts like partial derivatives and steepest ascent, we understand how the function behaves near specific points. Optimization combines these concepts to make calculated decisions about where the function reaches its desired value.
Applications of function optimization abound in fields like economics, engineering, and operations research, where optimal solutions can lead to cost savings, efficiency improvements, and strategic decision-making.