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Find the absolute maximum and minimum of the function \(f(x, y)=x^{2}+\) \(y^{2}\) subject to the constraint \(x^{4}+y^{4}=6561\). As usual, ignore unneeded answer blanks, and list points in lexicographic order. Absolute minimum value: ____________________. attained at ( ____________, ______________) ( ____________, ) Absolute minimum value: ____________________. attained at ( ____________, ______________) ( ____________, )

Short Answer

Expert verified
Absolute minimum value: 81. attained at (-9, 0), (0, -9) Absolute minimum value: 81. attained at (9, 0), (0, 9)

Step by step solution

01

Set up the Lagrange function

The Lagrange function is given by \(L(x, y, \lambda) = f(x, y) - \lambda g(x, y)\), where g(x, y) is the constraint function. In our case, f(x, y) = \(x^2 + y^2\) and g(x, y) = \(x^4 + y^4 - 6561\). So, the Lagrange function is: \(L(x, y, \lambda) = x^2 + y^2 - \lambda (x^4 + y^4 - 6561)\).
02

Compute the gradient of f(x, y) and the constraint function g(x, y)

We need to compute the gradient, which is the partial derivatives with respect to x, y, and \(\lambda\). So we have the following partial derivatives: \(\frac{\partial L}{\partial x} = 2x - \lambda 4x^3 = 0\) \(\frac{\partial L}{\partial y} = 2y - \lambda 4y^3 = 0\) \(\frac{\partial L}{\partial \lambda} = x^4 + y^4 - 6561 = 0\)
03

Equate the gradients and solve for x and y

From the partial derivatives with respect to x and y, we get two equations in x and y: \(2x - \lambda 4x^3 = 0 \Rightarrow x(1 - 2\lambda x^2) = 0\) \(2y - \lambda 4y^3 = 0 \Rightarrow y(1 - 2\lambda y^2) = 0\) Also, we have the constraint equation: \(x^4 + y^4 = 6561\).
04

Find the critical points (x, y)

Either x = 0 or \(1 - 2\lambda x^2 = 0\). Either y = 0 or \(1 - 2\lambda y^2 = 0\). Case 1: (x = 0, y ≠ 0) – In this case, when plugging x = 0 into constraint equazione, we get \(y^4 = 6561\). Taking the fourth root, we have y = ±9, x = 0. (0, -9) and (0, 9) are critical points. Case 2: (x ≠ 0, y = 0) – Similar to Case 1. The critical points are (-9, 0) and (9, 0). Case 3: Both x and y are non-zero. We have \(1 - 2\lambda x^2 = 0\) and \(1 - 2\lambda y^2 = 0\). We can solve these equations to find that \(x^2 = y^2\). But, we also have the constraint equation \(x^4 + y^4 = 6561\). Combining the two equations, we find that there are no solutions satisfying both equations. So the critical points are (-9, 0), (9, 0), (0, -9), and (0, 9).
05

Evaluate f(x, y) at the critical points

Now we evaluate the function \(f(x,y) = x^2 + y^2\) at these critical points: f(-9, 0) = (-9)^2 + 0 = 81 f(9, 0) = 9^2 + 0 = 81 f(0, -9) = 0 + (-9)^2 = 81 f(0, 9) = 0 + 9^2 = 81 Since all the values of the function at the critical points are equal, we have: Absolute minimum value: 81, attained at (-9, 0) and (0, -9) Absolute maximum value: 81, attained at (9, 0) and (0, 9)

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Key Concepts

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

Multivariable Calculus
Multivariable calculus is an extension of calculus that deals with functions of multiple variables, such as two or more, rather than just one. In this context, we're considering the function \( f(x, y) = x^2 + y^2 \). By understanding how this function behaves as both \(x\) and \(y\) change, we can explore a range of features like slopes, rates of change, and extrema (or extreme values).
When dealing with multivariable functions, one key element is understanding the concept of partial derivatives. These derivatives help analyze how a function changes with respect to one variable at a time, keeping the others constant. Here, in our exercise, when we compute gradients, we're essentially looking at these partial derivatives.
This foundational knowledge of multivariable calculus is crucial for tackling more complex topics such as optimization and constraints, as it provides the necessary tools to understand changes in quantity across multiple dimensions.
Optimization
Optimization is all about finding the maximum or minimum values of a function. In other words, it's like hunting for the 'best' or 'worst' possible outcomes that a function can give, based on certain conditions.
There are two main types of optimization: unconstrained and constrained. Unconstrained optimization looks for maxima or minima without any restrictions. Constrained optimization, like in our problem here, finds these extremes but within given limitations.
In the problem we've got, we're asked to find the extreme values of a function \( f(x, y) = x^2 + y^2 \). But we're bound by a constraint: \( x^4 + y^4 = 6561 \). This means, our task is not just to find the maxima and minima but to do so while adhering to this specific condition.
That's where the method of Lagrange multipliers comes into play. It's a nifty tool that helps in dealing with such constraints in optimization problems, ensuring that we don't step outside the bounds of the constraints set for us.
Constraint
Constraints are the limits or conditions placed upon the values that variables in a function can assume, defining the region of interest for our solutions. In the exercise we're tackling, the constraint is given by \( x^4 + y^4 = 6561 \).
Such constraints are crucial because they restrict the domain and guide where the function's extreme values are to be found. Constraints can be in different forms, such as equations like we see here, or inequalities which are quite common in other problems.
These constraints play a role not only in practical applications, like maximizing profit or minimizing cost in a real-world scenario, but they are also essential in ensuring that the solutions to the mathematical problems are realistic and applicable. In mathematical optimization, respecting constraints is necessary to find valid solutions that comply with any given conditions or limitations.
Critical Points
Critical points are specific points on a function where the first derivative (in each variable) is zero or undefined. In multivariable calculus, these are the points we scrutinize to determine potential extreme values—either maximums, minimums, or saddle points.
For the given function \( f(x, y) = x^2 + y^2 \) and the constraint \( x^4 + y^4 = 6561 \), we seek critical points by setting up the Lagrange function and finding where the gradients of our function and constraint are parallel.
Here, we calculated the critical points as \((-9, 0)\), \((9, 0)\), \((0, -9)\), and \((0, 9)\). After examining the function values at these points, we determined all yield the same value, meaning the extremes coincide at each of these points.
Finding critical points is a way to zero in on where changes happen in the function and where those bound by conditions give the desired extremes. This involves analyzing the nature of these points and verifying if they are indeed extremal (highest or lowest) given the context.

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Most popular questions from this chapter

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