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Of all rectangles with area 196, which one has the minimum perimeter? Let P and \( w \) be the perimeter and width, respectively, of the rectangle. Write the objective function in terms of P and w. Assume that the width is less than the length if the dimensions are unequal. \( \mathrm{P}=\square \) (Type an expression.) The interval of interest of the objective function is \( \square \). (Simplify your answer. Type your answer in interval notation.) Of all rectangles with area 196, the one with dimensions \( \square \) has the minimum perimeter. (Simplify your answers. Use a comma to separate answers as needed.)

Ask by Barber Cummings. in the United States
Feb 04,2025

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The rectangle with the minimum perimeter is a square with dimensions 14 by 14.

Solución

To determine the rectangle with the minimum perimeter among all rectangles with an area of 196, we'll follow these steps: 1. **Express the Perimeter \( P \) in terms of the Width \( w \):** Given: - Area \( A = 196 \) - \( A = l \times w \) where \( l \) is the length. Solving for \( l \): \[ l = \frac{196}{w} \] The perimeter \( P \) of a rectangle is: \[ P = 2(l + w) \] Substituting \( l \): \[ P = 2\left(\frac{196}{w} + w\right) = 2w + \frac{392}{w} \] **Objective Function:** \[ \boxed{P = 2w + \frac{392}{w}} \] 2. **Determine the Interval of Interest for \( w \):** - The width \( w \) must be positive: \( w > 0 \). - To ensure \( w \leq l \) (since \( w < l \) if dimensions are unequal), and given \( l = \frac{196}{w} \), the maximum value of \( w \) occurs when \( l = w \), i.e., when the rectangle becomes a square. - Solving \( w = \frac{196}{w} \) gives \( w = 14 \). **Interval of Interest:** \[ \boxed{(0, \ 14]} \] 3. **Identify the Dimensions that Minimize the Perimeter:** - The perimeter is minimized when the rectangle is a square. - From the area \( 196 \), each side of the square is: \[ w = l = \sqrt{196} = 14 \] **Dimensions with Minimum Perimeter:** \[ \boxed{14,\ 14} \]

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To find the perimeter \( P \) of a rectangle with fixed area 196, we can express the perimeter in terms of the width \( w \). The area \( A \) is given by \( A = w \cdot l \), where \( l \) is the length. Therefore, \( l = \frac{196}{w} \). The perimeter \( P \) can be expressed as: \[ P = 2w + 2l = 2w + 2\left(\frac{196}{w}\right) = 2w + \frac{392}{w} \] The interval of interest for the width \( w \) is determined by the condition that \( w \) must be positive and less than the length. Since the area \( A = 196 \), we have: \[ w > 0 \quad \text{and} \quad w < \frac{196}{w} \] This simplifies to \( w^2 < 196 \), so: \[ w < 14 \] Thus, the interval of interest can be expressed in interval notation as \( (0, 14) \). To minimize the perimeter, we set the derivative of \( P \) with respect to \( w \) to zero, which gives us \( w = 14 \) as the critical point. However, since the problem requires us to keep the width less than the length, we know from the area constraint that the rectangle with minimum perimeter occurs when \( w \) approaches 14 as closely as possible while remaining valid within the interval. Thus, the rectangle with dimensions where this perimeter is minimized is when: \[ w = 14, \quad l = \frac{196}{14} = 14 \] Ultimately, the rectangle that minimizes the perimeter has equal dimensions: Of all rectangles with area 196, the one with dimensions \( 14, 14 \) has the minimum perimeter.

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