# Lagrang multipliers to find max and min

• naspek
In summary, using the method of Lagrange multipliers, the highest and lowest temperatures encountered by an ant walking on a metal plate with temperature T(x, y) = 4x^2 − 4xy + y^2 are (5^1/2, 5^1/2) and (5^1/2, -(5^1/2)) respectively. These critical points are obtained by substituting the given constraint x^2 + y^2 = 5^2 into the equation 8x - 4y = 2x and 2y - 4x = 2y.
naspek

## Homework Statement

The temperature at a point (x, y) on a metal plate is T(x, y) = 4x^2 − 4xy + y^2 .
An ant, walking on the plate, traverses a circle of radius 5 centered at the origin.
Using the method of Lagrange multipliers, find the highest and lowest
temperatures encountered by the ant.

## The Attempt at a Solution

T(x,y) = 4x^2 − 4xy + y^2
gradient of T = (8x - 4y)i + (2y - 4x)j

g(x,y) = x^2 + y^2 = 5^2
gradient of g = (2x)i + (2y)j

8x - 4y = #2x ---->1
2y - 4x = #2y ---->2

# = 4-4y = 1-4x

what am i going to do next?

haven't checked your work, but notice you also have the original constraint, use that with the equation that you found

lanedance said:
haven't checked your work, but notice you also have the original constraint, use that with the equation that you found

what is the meaning of original constraint?

naspek said:
what is the meaning of original constraint?

you have a function to optimise T(x, y) = 4x^2 − 4xy + y^2, against a given constraint x^2 + y^2 = 5^2

if i solved the previous equation, i get, x=(1/4)y

i've got max= 45 and min=5..
am i done it right?

#### Attachments

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Yes, what you have done so far is correct. But that certainly does NOT give "max= 45 and min=5"!

ok...
when y = 2x,
i substitute into x^2+y^2=25
i got x=+-(5)^1/2

when x=-2y,
i substitute into x^2+y^2=25
i got y= +-(5)^1/2

so.. i got my critical points..
---> [(5^1/2) , (5^1/2)]
---> [(5^1/2) , -(5^1/2)]
---> [-(5^1/2) , (5^1/2)]
---> [-(5^1/2) , -(5^1/2)]

my critical points are correct?

## 1. What is the Lagrangian function and how does it relate to finding max and min?

The Lagrangian function is a mathematical tool used in optimization problems to find the maximum or minimum value of a function subject to constraints. It is used in conjunction with Lagrange multipliers to find the extreme values of a multivariable function.

## 2. How do Lagrange multipliers work?

Lagrange multipliers are a method of finding the extreme values of a function subject to constraints by introducing additional variables called multipliers. These multipliers are used to convert the constrained optimization problem into an unconstrained one, making it easier to find the extreme values.

## 3. When should Lagrange multipliers be used?

Lagrange multipliers are most commonly used when trying to find the maximum or minimum value of a function subject to constraints. They are particularly useful in cases where it is difficult to directly solve for the extreme values using traditional methods.

## 4. What are the limitations of using Lagrange multipliers?

One limitation of using Lagrange multipliers is that they are only applicable to differentiable functions. Additionally, they may not always provide the global maximum or minimum of a function, but rather a local one. Therefore, it is important to check the results obtained using Lagrange multipliers to ensure they are the desired extreme values.

## 5. Can Lagrange multipliers be used in real-world applications?

Yes, Lagrange multipliers are commonly used in various fields of science and engineering to solve optimization problems. They are particularly useful in economics, physics, and engineering to find the maximum or minimum value of a function subject to constraints in real-world scenarios.

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