Chapter 4: Problem 2
A rod of fixed length \(k\) slides along the coordinate axes. If it meets the axes at \(A(a, 0)\) and \(B(0, b)\), then the minimum value of \(\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}\) is a. 0 b. 8 c. \(k^{2}-4+\frac{4}{k^{2}}\) d. \(k^{2}+4+\frac{4}{k^{2}}\)
Short Answer
Expert verified
The minimum value of the expression is 8, so the answer is b.
Step by step solution
01
Understand the Problem
We are given a rod of length \(k\) that meets the coordinate axes at points \(A(a, 0)\) and \(B(0, b)\). We need to find the minimum value of the expression \(\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}\) given the constraint \(a^2 + b^2 = k^2\).
02
Use Constraint to Relate Variables
Given that the rod length is constant, we have \(a^2 + b^2 = k^2\). We can use this relationship to express one variable in terms of the other, for example, \(b^2 = k^2 - a^2\).
03
Substitute and Simplify
Substitute \(b\) in the expression \(\left(b+\frac{1}{b}\right)^{2}\): We have \(b = \sqrt{k^2 - a^2}\) and \(b + \frac{1}{b}\), where \(b eq 0\). Substitute in the expression to get \(\left(\sqrt{k^2-a^2} + \frac{1}{\sqrt{k^2-a^2}}\right)^2\).
04
Use AM-GM Inequality
By the AM-GM inequality, we know \((a + \frac{1}{a}) \geq 2\) and \((b + \frac{1}{b}) \geq 2\). This means \(\left(a+\frac{1}{a}\right)^2 + \left(b+\frac{1}{b}\right)^2 \geq 2^2 + 2^2 = 8\).
05
Determine if Minimum is Achievable
The minimum value of 8 is achievable when both \((a + \frac{1}{a}) = 2\) and \((b + \frac{1}{b}) = 2\). This implies \(a = 1\) and \(b = 1\) since \(a + \frac{1}{a} = 2 \implies a = 1\). Therefore, the rod's length condition \(a^2 + b^2 = k^2\) becomes \(1^2 + 1^2 = 2 = k^2\). Hence, the minimum value is realized when \(k = \sqrt{2}\). Finally, the minimum value of the expression is 8.
06
Choose the Correct Answer
Comparing the solution to the provided options, 8 matches option b. Thus, the minimum value of the given expression is 8.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rod Length Constraint
In the given problem, we have a rod with a fixed length, denoted by the constant variable \(k\). This rod slides along the x and y axes, forming endpoints on the axes, specifically at point \(A(a, 0)\) on the x-axis, and point \(B(0, b)\) on the y-axis. The length of the rod introduces a crucial constraint due to the Pythagorean theorem from geometry:
- The coordinates \((a, 0)\) and \((0, b)\) are connected by the rod, resulting in the equation \(a^2 + b^2 = k^2\).
AM-GM Inequality
The Arithmetic Mean-Geometric Mean (AM-GM) Inequality is a pivotal inequality in mathematics, and it's highly useful for optimization and minimization tasks. It states that for any non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. For any non-zero real number \(x\), this means:
- \((x + \frac{1}{x}) \geq 2\), with equality when \(x = 1\).
Minimization Problem
In mathematics, solving a minimization problem involves finding a value that makes a given function reach its lowest point. Here, our focus is on determining the minimum value of the expression:
- \(\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}\).
Coordinate Axes Intersection
The role of coordinate axes intersection in this problem gives a geometric visualization to the given equation. The rod, intersecting the x-axis and y-axis at points \((a, 0)\) and \((0, b)\) respectively, serves as a visual cue for the application of algebraic principles. These intersections essentially translate a theoretical algebraic concept into a tangible spatial scenario:
- The points \((a,0)\) and \((0,b)\) highlight the maximum extent on x and y, respectively, within which the rod can remain under a given constraint.
- From understanding this intersection, it’s easier to align geometric concepts (like the Pythagorean theorem) with algebraic constraints.