Vector Algebra
Dot Product and Optimization
Grade None
Question:
<p>Let <span class="math">\mathbf{a}, \mathbf{b} > 0</span> and <span class="math">\boldsymbol{\alpha} = \frac{4}{a}\mathbf{i} + \frac{j}{b} + \mathbf{b}\mathbf{k}</span> and <span class="math">\boldsymbol{\beta} = \mathbf{b}\mathbf{i} + \mathbf{a}\mathbf{j} + \frac{1}{b}\mathbf{k}</span>, then the maximum value of <span class="math">\frac{10}{5 + \boldsymbol{\alpha} \cdot \boldsymbol{\beta}}</span> is</p>
<p>(a) 1</p>
<p>(b) 2</p>
<p>(c) 4</p>
<p>(d) 8</p>
Step-by-Step Solution
Key Concept: Use AM-GM inequality to find the minimum value of the dot product, then compute the maximum of the reciprocal expression.
Solution: Computing the dot product: \boldsymbol{\alpha} \cdot \boldsymbol{\beta} = \frac{4}{a} \cdot \mathbf{b} + \frac{1}{b} \cdot \mathbf{a} + \mathbf{b} \cdot \frac{1}{b} = \frac{4b}{a} + \frac{a}{b} + 1 By AM-GM inequality: \frac{4b}{a} + \frac{a}{b} \geq 2\sqrt{\frac{4b}{a} \cdot \frac{a}{b}} = 2\sqrt{4} = 4 \therefore \boldsymbol{\alpha} \cdot \boldsymbol{\beta} \geq 4 + 1 = 5 Therefore: \frac{10}{5 + \boldsymbol{\alpha} \cdot \boldsymbol{\beta}} \leq \frac{10}{5 + 5} = \frac{10}{10} = 1 The maximum value is 1.
Correct Answer: A