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Chapter 4: Problem 8

Of all rectangles with a fixed area \(A,\) which one has the minimum perimeter? (Give the dimensions in terms of \(A .\) )

Short Answer

Expert verified
Answer: The dimensions of the rectangle are both equal to the square root of the fixed area (A). That is, the rectangle is a square with side lengths \(\sqrt{A} \times \sqrt{A}\).

Step by step solution

01

Introduce Variables and Constraints

Let's introduce variables for length (l) and width (w) of the rectangle. Also, we are given that the area of the rectangle A is fixed, so we have the constraint: \[lw = A\]
02

Express Perimeter in Terms of Width and Area

The perimeter of a rectangle is given by the sum of twice the length and twice the width, or P = 2l + 2w. We can express the perimeter in terms of width and area using the constraint from Step 1: \[P(w) = 2w + 2\frac{A}{w}\]
03

Find the Derivative of the Perimeter

In order to optimize the perimeter function, we need to find its derivative with respect to w. We will then find the critical points, where the derivative is 0 or undefined. These points can be local minima, local maxima, or inflection points. \[P'(w) = \frac{dP(w)}{dw}\] We can find the derivative using the power rule and the chain rule: \[P'(w) = 2 - \frac{2A}{w^2}\]
04

Find the Critical Point

Now, we want to find the critical points by setting the derivative to 0 and solving for w: \[P'(w)=0 \Rightarrow 2 - \frac{2A}{w^2} = 0\] Solving for w, we find the critical point: \[w^2 = A \Rightarrow w = \sqrt{A}\]
05

Determine the dimensions of the rectangle

Now that we have the width, we can plug it back into our constraint equation from Step 1 to determine the length: \[l\sqrt{A} = A\] Solving for l, we get: \[l = \frac{A}{\sqrt{A}} = \sqrt{A}\] So, the dimensions of the rectangle with a fixed area A that has the minimum perimeter are both equal to the square root of A. In other words, the rectangle is a square with side length \(\sqrt{A}\). Dimensions: \(\sqrt{A} \times \sqrt{A}\)

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