Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 13 - Partial Derivatives - 13.9 Lagrange Multipliers - Exercises Set 13.9 - Page 997: 36

Answer

See explanation

Work Step by Step

Step 1: Formulating the Problem Using the perimeter constraint \(2\ell + 2w = p\) where \(0 \leq \ell, w\) and \(0 \leq x, y\), we get: \[ \begin{align*} \ell &= \frac{p}{2} - x \\ w &= \frac{p}{2} - y \end{align*} \] So, \[ A = \ell \cdot w = \left(\frac{p}{2} - x\right) \cdot \left(\frac{p}{2} - y\right) = \frac{p}{4} \cdot \left(\frac{p}{2} - x\right) \cdot \left(\frac{p}{2} - y\right) \] Step 2: Since \(\ell\) and \(w\) represent lengths, they can't be negative, and the perimeter shouldn't exceed \(p\), so: \[ 0 \leq \ell \leq \frac{p}{2} \quad \text{and} \quad 0 \leq w \leq \frac{p}{2} \] \(A\) is a continuous function on \([0, \frac{p}{2}]\), so the maximum occurs at an endpoint of this interval or at a critical point. Step 3: Finding Critical Point The critical point fulfills: \[ \frac{dA}{dx} = 0 \quad \text{and} \quad \frac{dA}{dy} = 0 \] \[ \frac{dp}{dx} = 0 \quad \text{and} \quad \frac{dp}{dy} = 0 \] So, \[ x = \frac{p}{4} \quad \text{and} \quad y = \frac{p}{4} \] So the maximum occurs at \((x, y) = \left(\frac{p}{4}, \frac{p}{4}\right)\). Step 4: Substituting into \(A\) Substituting into \(A = \frac{p}{4} \cdot \left(\frac{p}{2} - x\right) \cdot \left(\frac{p}{2} - y\right)\), we get: For \(x = 0\): \[ A = 0 \] For \(x = \frac{p}{4}\): \[ A = \frac{p}{4} \cdot \left(\frac{p}{2} - \frac{p}{4}\right) \cdot \left(\frac{p}{2} - \frac{p}{4}\right) = \frac{p^2}{16} \] For \(x = \frac{p}{2}\): \[ A = \frac{p}{4} \cdot \left(\frac{p}{2} - \frac{p}{2}\right) \cdot \left(\frac{p}{2} - \frac{p}{2}\right) = 0 \] So, the maximum occurs at \((x, y) = \left(\frac{p}{4}, \frac{p}{4}\right)\) and is equal to \(\frac{p^2}{16}\). Step 5: Comparison with Example 2 We obtained the same results as in Example 2. However, there are differences in the methods. In this approach, we used the perimeter constraint as a starting point and expressed \(y\) in terms of \(x\) to find a one-variable expression of \(A\). In Example 2, we only used the constraint to solve Lagrange multipliers, which gave us a shortcut to the solution (by realizing that \(\ell = w = x = y\)).
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.