Problem 56 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Deal with topics developed in Chapter \(7 .\) Find the largest and the smallest values of the function $$ f(x, y)=2 \sin x+5 \cos y $$ over the rectangle \(R\) with vertices \((0,0),(2,0)\), \((2,5)\), and \((0,5)\).

Step-by-Step Solution

Answer

largest value=5 smallest =-5

Step 1 - Identify the function and region

The given function is \[f(x, y) = 2 \sin x + 5 \cos y\]and the region is the rectangle \(R\) with vertices \((0,0), (2,0), (2,5), (0,5)\).

Step 2 - Evaluate the function at the corners of the rectangle

Evaluate \(f(x, y)\) at each corner: \(f(0,0) = 2 \sin 0 + 5 \cos 0 = 2 \cdot 0 + 5 \cdot 1 = 5\)\(f(2,0) = 2 \sin 2 + 5 \cos 0 = 2 \sin 2 + 5\)\(f(2,5) = 2 \sin 2 + 5 \cos 5\)\(f(0,5) = 2 \sin 0 + 5 \cos 5 = 5 \cos 5\)

Step 3 - Evaluate the function along the edges of the rectangle

Evaluate \(f(x, y)\) along each edge. For the edge from \((0,0)\) to \((2,0)\), the function is evaluated at \(y = 0\): \(f(x,0) = 2 \sin x + 5\)For the edge from \((2,0)\) to \((2,5)\), the function is evaluated at \(x = 2\): \(f(2,y) = 2 \sin 2 + 5 \cos y\)For the edge from \((2,5)\) to \((0,5)\), the function is evaluated at \(y = 5\): \(f(x, 5) = 2 \sin x + 5 \cos 5\)For the edge from \((0,5)\) to \((0,0)\), the function is evaluated at \(x = 0\): \(f(0,y) = 2 \sin 0 + 5 \cos y = 5 \cos y\)

Step 4 - Determine the maximum and minimum values

Compare the values found:a) \(f(0,0) = 5\)b) \(f(2,0) = 2 \sin 2 + 5\)c) \(f(2, 5) = 2 \sin 2 + 5 \cos 5\)d) \(f(0, 5) = 5 \cos 5 - 5\) e) Along the edges we find f(0,y) evaluates 5 cos(y). From which max value is 5 and min is -5.

Key Concepts

partial derivatives

Partial derivatives allow us to find the rate at which a function is changing with respect to one of its variables, while keeping the other variables constant. In the context of multivariable functions like our given function \[f(x, y) = 2 \sin x + 5 \cos y\], partial derivatives help determine the slope of the function along the x and y directions separately.

The partial derivative with respect to \(x\), denoted as \(\frac{\partial f}{\partial x}\), measures the change in \(f\) when only \(x\) is varied:
\[\frac{\partial f}{\partial x} = \frac{d}{dx}(2 \sin x + 5 \cos y) = 2 \cos x\]
The partial derivative with respect to \(y\), denoted as \(\frac{\partial f}{\partial y}\), measures the change in \(f\) when only \(y\) is varied:
\[\frac{\partial f}{\partial y} = \frac{d}{dy}(2 \sin x + 5 \cos y) = -5 \sin y\]

To find critical points where extrema might occur, we set these partial derivatives to zero and solve for \(x\) and \(y\):

• \(2 \cos x = 0\)
• \(-5 \sin y = 0\)

Solving these equations:

• \(\cos x = 0 \Rightarrow x = \frac{\pi}{2} + n\pi, \: \: n \in \mathbb{Z}\)
• \(\sin y = 0 \Rightarrow y = n\pi, \: \: n \in \mathbb{Z}\)

By evaluating these solutions within our given region, we can then identify potential maxima and minima points.

multivariable functions

Multivariable functions are functions that have more than one input variable. For example, the function \[f(x, y) = 2 \sin x + 5 \cos y\] depends on both \(x\) and \(y\).

These functions can be visualized as surfaces in three-dimensional space, where each point on the surface represents a value of the function for specific values of \(x\) and \(y\). Understanding how these functions behave in different regions requires tools like partial derivatives and boundary evaluations.

In this exercise, we’re interested in finding the maximum and minimum values of the function within a defined region. This involves:

• Evaluating the function at specified points (e.g., corners of the region)
• Analyzing variations along the edges
• Using optimization techniques like the Extreme Value Theorem

Grasping the behavior of multivariable functions is crucial for various fields such as physics, engineering, and economics, where many phenomena depend on multiple factors simultaneously.

boundary evaluation

Boundary evaluation is a critical step in solving extremum problems for multivariable functions within a restricted area, such as a rectangle. In our problem with the function \[f(x, y) = 2 \sin x + 5 \cos y\], we need to evaluate the function along the edges of the rectangle defined by the vertices \((0,0), (2,0), (2,5), (0,5)\).

To perform boundary evaluation, we:

• Assess the function at the corner points
• Evaluate the function along each edge:

• **Edge (0,0) to (2,0):** Along \(y = 0\), the function is \[f(x,0) = 2 \, \sin x + 5\]
• **Edge (2,0) to (2,5):** Along \(x = 2\), the function is \[f(2,y) = 2 \sin 2 + 5 \cos y\]
• **Edge (2,5) to (0,5):** Along \(y = 5\), the function is \[f(x, 5) = 2 \sin x + 5 \cos 5\]
• **Edge (0,5) to (0,0):** Along \(x = 0\), the function is \[f(0,y) = 5 \cos y\]

By comparing values at these points and analyzing the edge functions, we can determine the maximum and minimum values within the rectangle.

Boundary evaluation ensures no potential extremum is missed within the region’s boundaries, offering a comprehensive analysis.