Problem 35
Question
Find the maximum and minimum values of \(f(x, y)=10+x+y\) on the disk \(x^{2}+y^{2} \leq 9 .\) Hint: Parametrize the boundary by \(x=3 \cos t, y=3 \sin t, 0 \leq t \leq 2 \pi\)
Step-by-Step Solution
Verified Answer
Maximum is \(10 + 3\sqrt{2}\); minimum is \(10 - 3\sqrt{2}\).
1Step 1: Understand the Given Problem
We need to find the maximum and minimum values of the function \(f(x, y) = 10 + x + y\) within the disk defined by \(x^2 + y^2 \leq 9\). This involves checking both the interior of the disk and its boundary.
2Step 2: Evaluate at Critical Points Inside the Disk
Since \(f(x, y) = 10 + x + y\) is linear, the gradient \(abla f = (1, 1)\) is constant, and there are no critical points inside the circular region because \(f\) has no derivative (zero gradient) to analyze. We will directly evaluate the boundary.
3Step 3: Parametrize the Boundary
The boundary of the disk is a circle with radius 3. We use the given parameterization: \(x = 3 \cos t\), \(y = 3 \sin t\) with \(0 \leq t \leq 2\pi\). Substitute \(x\) and \(y\) into \(f\).
4Step 4: Substitute the Parametrization into the Function
Substitute the parametrized values into the function: \[f(x, y) = 10 + 3\cos t + 3\sin t\]. This simplifies to \(f(t) = 10 + 3(\cos t + \sin t)\).
5Step 5: Simplify the Function Using Trigonometric Identity
Use the identity \(\cos t + \sin t = \sqrt{2} \sin(t + \pi/4)\). This gives \(f(t) = 10 + 3\sqrt{2}\sin(t + \pi/4)\).
6Step 6: Find Extremes of the Simplified Function
The maximum value of \(\sin(t + \pi/4)\) is 1 and the minimum value is -1, giving the maximum of \(f(t) = 10 + 3\sqrt{2}\) and the minimum \(f(t) = 10 - 3\sqrt{2}\).
7Step 7: Consider the Interior (Verification Step)
Review to ensure no overlooked points; since \(f(x, y)\) is linear, no extremas will be in the interior due to the monotonic increasing nature along any line direction.
8Step 8: Conclude with Maximum and Minimum Values
The maximum value of the function on the disk is \(10 + 3\sqrt{2}\) and the minimum value is \(10 - 3\sqrt{2}\).
Key Concepts
OptimizationParametrizationGradientTrigonometric Identities
Optimization
Optimization is a key concept in multivariable calculus. It involves finding the maximum or minimum values of a function, usually subject to certain constraints. In the exercise, the function is given by \( f(x, y) = 10 + x + y \) on a disk \( x^2 + y^2 \leq 9 \). To solve for optimization, you need to examine both the interior of the region and its boundary.
In cases of circular boundaries, as given in this problem, one method is to first check for critical points within the region. Here, because the function is linear, \( abla f = (1, 1) \), the gradient does not change, indicating no critical points inside the disk. Hence, we focus on the boundary to determine optimization values.
This particular problem highlights the importance of moving your focus to the boundary when the interior does not yield any critical points, a common scenario in optimization problems involving linear functions.
In cases of circular boundaries, as given in this problem, one method is to first check for critical points within the region. Here, because the function is linear, \( abla f = (1, 1) \), the gradient does not change, indicating no critical points inside the disk. Hence, we focus on the boundary to determine optimization values.
This particular problem highlights the importance of moving your focus to the boundary when the interior does not yield any critical points, a common scenario in optimization problems involving linear functions.
Parametrization
Parametrization helps us transform a problem from one set of variables to another. When dealing with boundary problems in a circular region, converting to parametric form can simplify calculations.
In our exercise, the boundary of the disk is expressed using parametric equations: \( x = 3 \cos t \) and \( y = 3 \sin t \), with \( 0 \leq t \leq 2 \pi \). Parametrization is crucial in this context because it transforms the circle into a single variable problem, making it easier to manage. Instead of calculating over a two-dimensional domain \( x \) and \( y \), you now work with the single parameter \( t \).
This approach allows for a streamlined way to substitute these into the function and focus only on changes caused by \( t \), which eventually leads to finding the optimized values along the circle's edge.
In our exercise, the boundary of the disk is expressed using parametric equations: \( x = 3 \cos t \) and \( y = 3 \sin t \), with \( 0 \leq t \leq 2 \pi \). Parametrization is crucial in this context because it transforms the circle into a single variable problem, making it easier to manage. Instead of calculating over a two-dimensional domain \( x \) and \( y \), you now work with the single parameter \( t \).
This approach allows for a streamlined way to substitute these into the function and focus only on changes caused by \( t \), which eventually leads to finding the optimized values along the circle's edge.
Gradient
The gradient of a function provides a direction of fastest increase or decrease and is fundamentally a vector. For the function given, \( f(x, y) = 10 + x + y \), the gradient \( abla f \) is simply \( (1, 1) \).
The constant gradient vector indicates that everywhere in the space, the function consistently increases at the same rate in the direction of \( (1, 1) \). Since no change in the gradient happens inside the circle \( x^2 + y^2 \leq 9 \), there aren’t any critical points within this region. Knowing this can save time in the optimization process, steering the focus merely on evaluating the function along specific boundaries.
In this exercise, recognizing that the gradient of \( f \) does not vary reveals why solutions may need to skew towards boundary evaluation rather than the region's interior.
The constant gradient vector indicates that everywhere in the space, the function consistently increases at the same rate in the direction of \( (1, 1) \). Since no change in the gradient happens inside the circle \( x^2 + y^2 \leq 9 \), there aren’t any critical points within this region. Knowing this can save time in the optimization process, steering the focus merely on evaluating the function along specific boundaries.
In this exercise, recognizing that the gradient of \( f \) does not vary reveals why solutions may need to skew towards boundary evaluation rather than the region's interior.
Trigonometric Identities
Trigonometric identities simplify expressions involving trigonometric functions, making calculations more manageable. In this exercise, the identity used is \( \cos t + \sin t = \sqrt{2} \sin(t + \pi/4) \).
Applying this identity simplifies the function \( f(t) = 10 + 3(\cos t + \sin t) \) to\( f(t) = 10 + 3\sqrt{2}\sin(t + \pi/4) \). The use of this identity is pivotal as it highlights the periodic nature of the function and allows us to easily identify maximum and minimum values by observing the sine function’s range: \([-1, 1]\).
Recognizing which trigonometric identities to use and applying them effectively can greatly ease the process of finding extreme values, especially in boundary problem scenarios like this where the parameter \( t \) is involved.
Applying this identity simplifies the function \( f(t) = 10 + 3(\cos t + \sin t) \) to\( f(t) = 10 + 3\sqrt{2}\sin(t + \pi/4) \). The use of this identity is pivotal as it highlights the periodic nature of the function and allows us to easily identify maximum and minimum values by observing the sine function’s range: \([-1, 1]\).
Recognizing which trigonometric identities to use and applying them effectively can greatly ease the process of finding extreme values, especially in boundary problem scenarios like this where the parameter \( t \) is involved.
Other exercises in this chapter
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