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Use Lagrange multipliers to find these values. $$f(x, y)=x^{2} \text { subject to } x^{2}+x y+y^{2}=3$$

Short Answer

Expert verified
Answer: The extreme values of the function $$f(x, y) = x^2$$, subject to the constraint $$x^2 + xy + y^2 = 3$$, are $$\frac{12}{5}$$. These values occur at the points $$(2\sqrt{\frac{3}{5}}, \sqrt{\frac{3}{5}})$$ and $$(-2\sqrt{\frac{3}{5}}, -\sqrt{\frac{3}{5}})$$.

Step by step solution

01

Set up the Lagrangian function

To solve this problem using Lagrange multipliers, we first need to define the Lagrangian function: $$L(x, y, \lambda) = f(x, y) - \lambda(g(x, y) - c),$$ Where $$\lambda$$ is the Lagrange multiplier, $$c$$ is the constant term in the constraint, and $$g(x, y)$$ is the constraint equation. In our case, we have $$f(x, y) = x^2$$ and $$g(x, y) = x^2 + xy + y^2 - 3$$. Thus, our Lagrangian function becomes: $$L(x, y, \lambda) = x^2 - \lambda(x^2 + xy + y^2 - 3)$$
02

Calculate the partial derivatives

Now, we need to calculate the partial derivatives with respect to $$x$$, $$y$$, and $$\lambda$$: $$\frac{\partial L}{\partial x} = 2x - 2\lambda x - \lambda y = 0,$$ $$\frac{\partial L}{\partial y} = -\lambda x + 2\lambda y = 0,$$ $$\frac{\partial L}{\partial \lambda} = x^2 + xy + y^2 - 3 = 0$$
03

Solve the system of equations

Now, we need to solve the system of equations formed by the partial derivatives. We have: $$2x(1 - \lambda) - \lambda y = 0 \text{ (1)}$$ $$-\lambda x + 2\lambda y = 0 \text{ (2)}$$ $$x^2 + xy + y^2 = 3 \text{ (3)}$$ Dividing (2) by $$\lambda$$, we obtain: $$-x + 2y = 0$$, or $$x = 2y$$. Plug this back into (1), we get: $$2(2y)(1 - \lambda) - \lambda y = 0 \Rightarrow \lambda = \frac{4y}{y+4}$$ Plug $$x=2y$$ and $$\lambda = \frac{4y}{y+4}$$ into (3), we obtain: $$5y^2 = 3 \Rightarrow y = \pm \sqrt{\frac{3}{5}}$$, and corresponding $$x$$ values are $$x = \pm 2\sqrt{\frac{3}{5}} $$
04

Find the extreme values

Now we plug the $$x$$ and $$y$$ values back into the function $$f(x, y)$$ to find the extreme values: $$f\left(2\sqrt{\frac{3}{5}}, \sqrt{\frac{3}{5}}\right) = f\left(-2\sqrt{\frac{3}{5}}, -\sqrt{\frac{3}{5}}\right) = \left(2\sqrt{\frac{3}{5}}\right)^2 = \frac{12}{5}$$ So, the extreme values of the function $$f(x, y) = x^2$$, subject to the constraint $$x^2 + xy + y^2 = 3$$, are $$\frac{12}{5}$$, which occurs at the points $$(2\sqrt{\frac{3}{5}}, \sqrt{\frac{3}{5}})$$ and $$(-2\sqrt{\frac{3}{5}}, -\sqrt{\frac{3}{5}})$$.

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Key Concepts

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

Constraint Optimization
In mathematics, constraint optimization involves optimizing a function subject to specific conditions or constraints. Here, we're interested in maximizing or minimizing a function with respect to these constraints. This is particularly useful in scenarios where certain conditions must be adhered to.
For the given exercise, the function we want to optimize is given by \( f(x, y) = x^2 \), and the constraint is \( x^2 + xy + y^2 = 3 \). The role of the constraint is to limit the set of possible solutions to those that fulfill the equation \( x^2 + xy + y^2 = 3 \).
Through constraint optimization, we can pinpoint where this function presents extreme values that are consistent with the constraint, thereby finding practical solutions in real-world situations like resource allocation or minimizing cost while meeting specific requirements.
Partial Derivatives
Partial derivatives are fundamental in the realm of calculus, particularly when dealing with functions of multiple variables. They express how a function changes as each of its variables is altered independently. In the context of Lagrange multipliers, partial derivatives help in determining where the function changes and how.
During our exercise, we calculated partial derivatives of the Lagrangian \( L(x, y, \lambda) = x^2 - \lambda(x^2 + xy + y^2 - 3) \) with respect to each variable: \( x \), \( y \), and the multiplier \( \lambda \). This step is crucial because it leads to the equations that define the conditions necessary for optimization under constraints.
By setting these derivatives equal to zero, we can derive a system of equations that form the basis for identifying the critical points of the given function with respect to its constraints.
Extreme Values
Extreme values, such as minimums and maximums, play a pivotal role in optimization problems. They reveal the points at which a function reaches its greatest or smallest value subject to certain conditions.
With the use of Lagrange multipliers, we determine these points by solving a system of equations derived from partial derivatives. In our exercise, once the system was solved, we substituted the values of \( x \) and \( y \) we found into the function \( f(x, y) = x^2 \) to calculate the extreme values.
Ultimately, we discovered that the function \( f(x, y) \) reaches its extreme value of \( \frac{12}{5} \) at the specific points \( (2\sqrt{\frac{3}{5}}, \sqrt{\frac{3}{5}}) \) and \( (-2\sqrt{\frac{3}{5}}, -\sqrt{\frac{3}{5}}) \). Understanding extreme values ensures that solutions are not only mathematically valid but also optimal.
System of Equations
A system of equations is a collection of equations with multiple variables that are solved simultaneously. They're central in solving optimization problems using Lagrange multipliers.
In our problem, the partial derivatives of the Lagrangian led to a system of three equations:
  • \( 2x(1 - \lambda) - \lambda y = 0 \)
  • \( -\lambda x + 2\lambda y = 0 \)
  • \( x^2 + xy + y^2 = 3 \)
We solved this system step by step:
  • First, dividing and rearranging these, to express variables like \( x \) or \( y \) in terms of each other (\( x = 2y \)).
  • Then substituting back into one of the original equations to find numerical values for variables like \( y \).
  • Finally, substituting these values into the constraint equation to ensure they satisfy the original constraint.
Such systems allow us to unravel complex optimization problems systematically, ensuring all conditions are respected.

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Most popular questions from this chapter

A snapshot (frozen in time) of a set of water waves is described by the function \(z=1+\sin (x-y),\) where \(z\) gives the height of the waves and \((x, y)\) are coordinates in the horizontal plane \(z=0\) a. Use a graphing utility to graph \(z=1+\sin (x-y)\) b. The crests and the troughs of the waves are aligned in the direction in which the height function has zero change. Find the direction in which the crests and troughs are aligned. c. If you were surfing on one of these waves and wanted the steepest descent from the crest to the trough, in which direction would you point your surfboard (given in terms of a unit vector in the \(x y\) -plane)? d. Check that your answers to parts (b) and (c) are consistent with the graph of part (a).

Distance from a plane to an ellipsoid (Adapted from 1938 Putnam Exam) Consider the ellipsoid \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1\) and the plane \(P\) given by \(A x+B y+C z+1=0 .\) Let \(h=\left(A^{2}+B^{2}+C^{2}\right)^{-1 / 2}\) and \(m=\left(a^{2} A^{2}+b^{2} B^{2}+c^{2} C^{2}\right)^{1 / 2}\) a. Find the equation of the plane tangent to the ellipsoid at the point \((p, q, r)\) b. Find the two points on the ellipsoid at which the tangent plane is parallel to \(P\), and find equations of the tangent planes. c. Show that the distance between the origin and the plane \(P\) is \(h\) d. Show that the distance between the origin and the tangent planes is \(h m\) e. Find a condition that guarantees the plane \(P\) does not intersect the ellipsoid.

Use Lagrange multipliers in the following problems. When the constraint curve is unbounded, explain why you have found an absolute maximum or minimum value. Box with minimum surface area Find the dimensions of the rectangular box with a volume of \(16 \mathrm{ft}^{3}\) that has minimum surface area.

Extreme points on flattened spheres The equation \(x^{2 n}+y^{2 n}+z^{2 n}=1,\) where \(n\) is a positive integer, describes a flattened sphere. Define the extreme points to be the points on the flattened sphere with a maximum distance from the origin. a. Find all the extreme points on the flattened sphere with \(n=2 .\) What is the distance between the extreme points and the origin? b. Find all the extreme points on the flattened sphere for integers \(n>2 .\) What is the distance between the extreme points and the origin? c. Give the location of the extreme points in the limit as \(n \rightarrow \infty\) What is the limiting distance between the extreme points and the origin as \(n \rightarrow \infty ?\)

Prove that the level curves of the plane \(a x+b y+c z=d\) are parallel lines in the \(x y\) -plane, provided \(a^{2}+b^{2} \neq 0\) and \(c \neq 0\).

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