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Find The Maximum And Minimum Values Of The Function F(x, Y) = 2x + 3y2 4x 5 On The Domain X2 + Y2 <

Find The Maximum And Minimum Values Of The Function F(x, Y) = 2x + 3y² - 4x + 5 On The Domain X² + Y² <

Understanding how to find the maximum and minimum values of a function within a specific domain is a fundamental concept in calculus and optimization. In this article, we will explore how to determine the maximum and minimum values of the function F(x, y) = 2x + 3y² - 4x + 5 on the domain defined by the inequality X² + Y² < r² (assuming the domain is the interior of a circle of radius r). This problem involves techniques such as critical point analysis and boundary examination, which are essential tools in multivariable calculus.

Introduction to the Problem

When working with functions of two variables, such as F(x, y), the goal is often to find the points within a specified domain where the function attains its highest and lowest values. These are known as the maximum and minimum, respectively. The domain in this case is the interior of a circle, defined by the inequality X² + Y² < r², where r is the radius.

The function in question is:

F(x, y) = 2x + 3y² - 4x + 5

which simplifies to:

F(x, y) = -2x + 3y² + 5

Now, our task is to find the maximum and minimum values of F(x, y) within the circle X² + Y² < r².

Step 1: Simplify the Function

Before proceeding, note that the function can be simplified:

    • F(x, y) = 2x + 3y² - 4x + 5
    • F(x, y) = (-2x) + 3y² + 5

This form makes it easier to analyze the behavior of the function.

Step 2: Find Critical Points Within the Domain

Critical points are points where the gradient of the function is zero. To find these points:

Partial Derivatives

Calculate the partial derivatives with respect to x and y:

    • ∂F/∂x = -2
    • ∂F/∂y = 6y

Set these derivatives equal to zero to find critical points:

    • ∂F/∂x = -2 ≠ 0 — The derivative with respect to x is constant and never zero.
    • ∂F/∂y = 6y = 0 ⇒ y = 0

Since ∂F/∂x is never zero, there are no critical points inside the domain based on the gradient being zero. However, the function's behavior along the boundary and at certain points should still be examined.

Step 3: Analyze the Boundary of the Domain

The boundary of the domain is the circle:

x² + y² = r²

Because the interior (x² + y² < r²) does not contain critical points, the maximum and minimum values are attained either at boundary points or at points approaching the boundary.

Parametrize the Boundary

Using parametric equations:

    • x = r cos θ
    • y = r sin θ

where θ varies from 0 to 2π.

Express F(x, y) on the Boundary

Substituting the parametric form into F(x, y):

F(r cos θ, r sin θ) = -2(r cos θ) + 3(r sin θ)² + 5

Simplify:

F(θ) = -2r cos θ + 3r² sin² θ + 5

Note that sin² θ = (1 - cos 2θ)/2, so:

F(θ) = -2r cos θ + 3r² (1 - cos 2θ)/2 + 5

which simplifies to:

F(θ) = -2r cos θ + (3r²/2) - (3r²/2) cos 2θ + 5

Step 4: Find the Extreme Values on the Boundary

To find the maximum and minimum of F(θ), analyze the function:

F(θ) = -2r cos θ - (3r²/2) cos 2θ + (3r²/2) + 5

Now, express cos 2θ in terms of cos θ:

Recall that cos 2θ = 2 cos² θ - 1. Thus:

F(θ) = -2r cos θ - (3r²/2)(2 cos² θ - 1) + (3r²/2) + 5

Distribute:

F(θ) = -2r cos θ - 3r² cos² θ + (3r²/2) + (3r²/2) + 5

Simplify:

F(θ) = -2r cos θ - 3r² cos² θ + 3r² + 5

Now, define x = cos θ, where x ∈ [-1, 1].

So, the problem reduces to analyzing the function:

G(x) = -2r x - 3r² x² + 3r² + 5, where x ∈ [-1, 1]

Finding Extreme Values of G(x)

Differentiate G(x) with respect to x:

dG/dx = -2r - 6r² x

Set derivative to zero to find critical points:

-2r - 6r² x = 0

Solve for x:

x = -2r / (6r²) = -2r / (6r²) = -1 / (3r)

Check whether this critical point x = -1 / (3r) lies within [-1, 1]:

    • If r > 1/3, then x = -1/(3r) ∈ (-1, 1), so the critical point is within domain.
    • If r ≤ 1/3, then x = -1/(3r) ≥ -1, but may be outside the domain depending on r.

Evaluating G(x) at Critical Point and Endpoints

Calculate G(x) at x = -1/(3r):

G(-1/(3r)) = -2r (-1/(3r)) - 3r² ( -1/(3r) )² + 3r² + 5

Simplify:


  • First term: -2r (-1/(3r)) = 2/3

  • Second term: -3r² (1 / (9 r²)) = -3r² (1 / (9 r²)) = -1/3

  • Sum: 2/3 - 1/3 + 3r² + 5 = (1/3) + 3r² + 5


Therefore:

G(-1/(3r)) = (1/3) + 3r² + 5 = (16/3) + 3r²

Now, evaluate G(x) at the endpoints x = -1 and x = 1:


  • For x = -1:


G(-1) = -2r (-1) - 3r² 1 + 3r² + 5 = 2r - 3r² + 3r² + 5 = 2r + 5

  • For x = 1:


G(1) = -2r 1 - 3r² 1 + 3r² + 5 = -2r - 3r² + 3r² + 5 = -2r + 5

Summary of boundary evaluations:

    • At x = -1: G = 2r + 5
    • At x = 1: G = -2r + 5
    • At critical point x = -1/(3r): G = (16/3) + 3r²

Step 5: Determine the Maximum and Minimum Values

Now, compare these values to find the maximum and minimum:


  • The maximum among the boundary points:

  • For r > 0, 2r + 5 increases with r, so the maximum at x = -1 is 2r + 5.

  • The critical point gives G = (16/3) + 3r², which increases with r².

  • The minimum:

  • For r > 0, -2r +

Frequently Asked Questions

What is the function we are analyzing for maximum and minimum values?
The function is F(x, y) = 2x + 3y^2 - 4x + 5.
What is the domain specified for the function F(x, y)?
The domain is the set of points where x^2 + y^2 < 1, i.e., the interior of the unit circle.
How do we find the critical points of the function within the domain?
By calculating the partial derivatives of F with respect to x and y, setting them equal to zero, and solving for x and y.
What are the partial derivatives of F(x, y)?
The partial derivatives are: ∂F/∂x = 2 - 4 and ∂F/∂y = 6y.
What do the critical points tell us about the maximum and minimum values?
Critical points indicate potential locations of local maxima or minima; these points need to be checked within the domain.
Are there any critical points inside the domain x^2 + y^2 < 1?
Yes, solving the derivatives shows that the critical point occurs at x = -1, y = 0, which lies outside the interior of the unit circle.
How do we evaluate the maximum and minimum values on the boundary of the domain?
By parametrizing the boundary x^2 + y^2 = 1 and evaluating F at those points to find extrema.
What is the parametrization of the boundary circle?
The boundary can be parametrized as x = cos θ, y = sin θ, where θ ranges from 0 to 2π.
How do we find the maximum and minimum values of F on the boundary?
Substitute x = cos θ and y = sin θ into F to get a function of θ, then analyze this function to find its maximum and minimum.
What is the final step to determine the maximum and minimum of F within the domain?
Compare the critical points' values (if any inside the domain) with the boundary evaluation results to identify the global maximum and minimum within x^2 + y^2 < 1.

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