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Chapter 9: Problem 9

If \(x^{2}+y^{2}=1\), find the least value of \(f(x, y)=x^{2}+y^{2}+x y+3 .\)

Short Answer

Expert verified
The minimum value of the function \(f(x, y)=x^{2}+y^{2}+x y+3\) given \(x^{2}+y^{2}=1\) is 4.

Step by step solution

01

Substituting into the function

Given the equation \(x^{2}+y^{2}=1\), this can be substituted into the function \(f(x, y)=x^{2}+y^{2}+x y+3 \). Doing this, we get \(f(x, y) = 1 + xy + 3 = xy + 4 .\)
02

Applying the arithmetic-geometric mean inequality

Now we apply the arithmetic-geometric mean (AM-GM) inequality in order to find the minimum value of \(f(x, y)\). The AM-GM inequality states that the arithmetic mean of a set of non-negative numbers is always greater than or equal to the geometric mean of those numbers. Using this inequality, we have: \(\frac{x+y}{2} \geq \sqrt{xy}.\) Solving for \(xy\) gives: \(xy \leq (\frac{x+y}{2})^2.\)
03

Substituting and finding the minimum value

Substitute the value of \(xy\) into the function as follows: \(f(x, y) = (\frac{x+y}{2})^2 + 4.\) The minimum value of \((\frac{x+y}{2})^2\) is 0 (which happens when \(x = -y\)), hence the minimum value of \(f(x, y)\) is 4.

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Key Concepts

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

Arithmetic-Geometric Mean Inequality
The Arithmetic-Geometric Mean Inequality.
is a powerful tool in mathematics used frequently to compare means.
The inequality states that for any set of non-negative numbers,
  • the arithmetic mean (average) is greater than or equal to the geometric mean.
In mathematical terms, for two non-negative numbers, the inequality can be expressed as:
\[\frac{x+y}{2} \geq \sqrt{xy}\]This can be extended to more than two numbers as well.
In the context of optimization problems,
  • it serves as a technique to find bounds or limits of certain expressions.
  • By comparing values derived from different types of means, you can make informed decisions about the minimum or maximum possible values.
Functions of Two Variables
Functions of two variables involve expressions where quantities depend on the values of two different variables.
An example of such a function is \(f(x, y) = x^2 + y^2 + xy + 3\),
  • which is a common situation in problems involving surfaces or planes in three-dimensional spaces.
  • Here, the variables \(x\) and \(y\) could represent different independent factors such as time and position.
When dealing with these functions, understanding the functional dependencies and their graphical interpretations is valuable.
You can visualize them as a surface in a 3D space, where evaluating the function at specific points \((x, y)\) will give corresponding heights or values on the surface.
By substituting known constraints into these functions, such as \(x^2 + y^2 = 1\), we simplify and reduce the complexity of the problem by focusing on essential aspects.
Optimization Problems
Optimization problems focus on finding the best or most efficient solution among several possibilities.
These include finding the minimum or maximum of a function within given constraints.
  • In the current example, finding the least value of the function \(f(x, y) = x^2 + y^2 + xy + 3\) given \(x^2 + y^2 = 1\) is an optimization problem.
The first step is often to substitute known relationships or constraints,
reducing the function to one with fewer variables or simpler terms, like in this exercise where it became \(xy + 4\).
By applying the Arithmetic-Geometric Mean Inequality, we can establish the bounds:
  • psiWe learned that the minimum value achieved when \( x = -y \).
Optimization techniques involve
  • taking these initial conditions and constraints into consideration,
  • and using methods like inequalities, calculus, or trial and error to find the desired solution.

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Most popular questions from this chapter

The circle \(x^{2}+y^{2}=1\) cuts the \(x\)-axis at \(P\) and \(Q\) Another circle with centre at \(Q\) and variable radius intersect the first circle at \(R\) above the \(\mathrm{x}\)-axis and line segment \(P Q\) at \(S\). Find the maximum area of the triangle \(Q S R .\)

Find the min value of \(x+y\), where \(x^{2}+y^{2}=1\)

Find the Min values of \(f(a)=a^{-5}+a^{-4}+3 a^{-3}+1+a^{8}+a^{10}, \mathrm{a}>0\)

Find the max value of \(y=\sqrt{x^{2}-1}+\sqrt{4-x^{2}}\).

The minimum radius vector of the curve \(\frac{a^{2}}{x^{2}}+\frac{b^{2}}{y^{2}}=1\) is of length (a) \((a-b)\) (b) \((a+b)\) (c) \(2(a+b)\) (d) \(2(a-b)\)
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