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Ant on a metal plate The temperature at a point \((x, y)\) on a metal plate is \(T(x, y)=4 x^{2}-4 x y+y^{2} .\) An ant on the plate walks around the circle of radius 5 centered at the origin. What are the highest and lowest temperatures encountered by the ant?

Short Answer

Expert verified
The highest and lowest temperatures are 25 and -75, respectively.

Step by step solution

01

Understand the Problem

The task is to find the highest and lowest values of \( T(x, y) = 4x^2 - 4xy + y^2 \) on the circle defined by \( x^2 + y^2 = 25 \). This requires us to find the extrema of a function within a constraint.
02

Apply the Method of Lagrange Multipliers

To find the extrema of \( T(x, y) \) under the constraint \( g(x, y) = x^2 + y^2 - 25 = 0 \), we use Lagrange multipliers. We need to set \( abla T = \lambda abla g \). The gradients are \( abla T = (8x - 4y, -4x + 2y) \) and \( abla g = (2x, 2y) \).
03

Solve the System of Equations

We have the system of equations: 1. \( 8x - 4y = 2\lambda x \)2. \( -4x + 2y = 2\lambda y \)3. \( x^2 + y^2 = 25 \)Solve these equations simultaneously.
04

Solve First Two Equations for \(\lambda\)

Dividing each first by \(x\) (assuming \(x eq 0\)) and the second by \(y\) (assuming \(y eq 0\)): From 1: \( \lambda = \frac{8 - 4y/x}{2} \) From 2: \( \lambda = \frac{-4x + 2y}{2y} \). Equating gives: \( 8 - 4y/x = -4x/y + 2 \).
05

Simplify and Find Specific Points

From the equation \( 8y = 4x \Rightarrow y = \frac{x}{2} \). Substitute \( y = \frac{x}{2} \) into \( x^2 + y^2 = 25 \) to get: \( x^2 + \left( \frac{x}{2} \right)^2 = 25 \Rightarrow \frac{5x^2}{4} = 25 \Rightarrow x^2 = 20 \Rightarrow x = \pm \sqrt{20} \Rightarrow x = \pm 2\sqrt{5} \). This gives \( y \) as \( \pm \sqrt{5} \).
06

Calculate Temperature at Extrema Points

Now calculate \( T(x, y) \) at points \( (2\sqrt{5}, \sqrt{5}) \): \( T = 4(4 \cdot 5) - 4(2\sqrt{5})(\sqrt{5}) + (\sqrt{5})^2 = 60 - 40 + 5 = 25 \)\( (-2\sqrt{5}, -\sqrt{5}) \): \( T = 60 - 40 + 5 = 25 \)\( (2\sqrt{5}, -\sqrt{5}) \): Calculate similarly to verify either symmetry or distinctness of different points.
07

Check and Identify the Extrema

Since \( T(2\sqrt{5}, \sqrt{5}) = T(-2\sqrt{5}, -\sqrt{5}) \), check other potential solutions from symmetry. Substitution yields consistent maximum at 25.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Extreme Values
When analyzing problems like the one involving the ant on a metal plate, the goal is to identify the extreme values of a function. These values can be either the highest (maximum) or lowest (minimum) temperatures encountered. Extreme values are crucial in understanding how a variable behaves within certain limits. In mathematical terms, these involve finding the values where a function reaches its peaks or troughs, particularly when there are constraints involved, such as the circular path in this exercise.

To determine these points, we often look for places where the derivative of the temperature function equals zero or is undefined, which indicates potential maximum or minimum values. However, when there is a constraint, the method of Lagrange multipliers becomes essential. This method helps find extreme values of a function which are restricted to lie on a given curve or surface. In our example, the temperature formula has to be evaluated while sticking to the circle defined by the equation \( x^2 + y^2 = 25 \).

Using this method, potential extreme value points were calculated, leading to temperatures found at points like \( (2\sqrt{5}, \sqrt{5}) \), showing an extreme temperature of 25. This aids in effectively pinpointing the maximum and minimum temperatures encountered by the ant on its constrained path.
Temperature Optimization
Temperature optimization is about determining the point where the temperature function reaches its best value (highest or lowest) within a certain region. In the ant problem, the optimization involves locating these temperature points while the ant is moving on the circle. The function \( T(x, y) = 4x^2 - 4xy + y^2 \) represents the temperature distribution across the metal plate, and the challenge is to find at what points on the circle \( x^2 + y^2 = 25 \) this function is maximized or minimized.

The optimization process begins by calculating derivatives, incorporating the restriction of the circle. We then examine the points where the gradient of the temperature function aligns with the gradient of the constraint, indicating potential optimal points. For instance, such a calculation resulted in finding the temperature at \( (2\sqrt{5}, \sqrt{5}) \) and similar symmetric points on the circle to verify the maximum temperature of 25.

This kind of temperature optimization is essential in practical applications such as designing heat-sensitive equipment or ensuring safety in thermal environments. It provides a precise method to predict and manage temperatures effectively along specified constraints.
Constraint Optimization
Constraint optimization is a powerful technique used to find the best solution of a function subject to certain restrictions or conditions. In the context of the ant on a metal plate, the constraint comes from the ant having to remain on a circle of radius 5. This essential condition transforms a simple optimization problem into a constraint optimization challenge.

We use the method of Lagrange multipliers to tackle such problems. This involves introducing an auxiliary variable, \( \lambda \) (a Lagrange multiplier), which helps manage the constraint alongside the function of interest. By setting the gradients of the temperature function and the constraint equal to each other, \( abla T = \lambda abla g \), we form a system of equations to solve for the optimal points.

In the original example, solving these equations leads to the points \( (\pm 2\sqrt{5}, \pm \sqrt{5}) \). These are tested to determine the extreme temperatures encountered. Constraint optimization ensures that while determining the best points, all conditions or limits set by the problem are respected, providing a robust solution framework for various practical scenarios.

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