2.10 Q-6

Question Statement

Find the lengths of the sides of a rectangle with an area of $36 , \text{cm}^2$ when its perimeter is minimum.

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Background and Explanation

To solve this problem, we need to find the dimensions of a rectangle that has a fixed area of 36 cm² and a minimum perimeter. We use the relationships between the area and perimeter of a rectangle, and apply calculus to minimize the perimeter by finding the critical points of the perimeter function.

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Solution

1. Define the variables:

   Let $x$ and $y$ represent the length and width of the rectangle, respectively (in cm). The area $A$ of the rectangle is given by:

$$A = x \cdot y = 36$$

Thus, we can express $y$ as:

$$y = \frac{36}{x} \quad \text{(Equation 1)}$$

2. Express the perimeter:

   The perimeter $p$ of the rectangle is given by:

$$p = 2x + 2y = 2\left(x + \frac{36}{x}\right)$$

This is the function we need to minimize.

3. Differentiate the perimeter function:

   To minimize the perimeter, we differentiate $p$ with respect to $x$:

$$\frac{dp}{dx} = 2 \left[ 1 + 36 \left( \frac{-1}{x^2} \right) \right]$$

Simplifying:

$$\frac{dp}{dx} = 2 \left( 1 - \frac{36}{x^2} \right) = 2 \left( \frac{x^2 - 36}{x^2} \right)$$

4. Set the derivative equal to zero:

   To find the critical points, set $\frac{dp}{dx} = 0$:

$$2 \left( \frac{x^2 - 36}{x^2} \right) = 0$$

Solving for $x$:

$$x^2 - 36 = 0 \quad \Rightarrow \quad x = 6$$

Since $x$ is positive, we take $x = 6$.

5. Check the second derivative:

   To confirm that this critical point corresponds to a minimum, compute the second derivative:

$$\frac{d^2p}{dx^2} = 2 \cdot \frac{d}{dx} \left( 1 - \frac{36}{x^2} \right)$$

Simplifying:

$$\frac{d^2p}{dx^2} = 2 \cdot \frac{72}{x^3}$$

Substituting $x = 6$:

$$\frac{d^2p}{dx^2} = \frac{144}{6^3} = \frac{2}{3}$$

Since the second derivative is positive, $x = 6$ corresponds to a minimum.

6. Find the dimensions of the rectangle:

   Using Equation 1, substitute $x = 6$ to find $y$:

$$y = \frac{36}{6} = 6$$

Therefore, the rectangle with area $36 , \text{cm}^2$ and minimum perimeter has dimensions $6 , \text{cm} \times 6 , \text{cm}$.

Thus, the side lengths of the rectangle are 6 cm and 6 cm.

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Key Formulas or Methods Used

• Area of a rectangle: $A = x \cdot y$
• Perimeter of a rectangle: $p = 2x + 2y$
• First derivative test: To find the critical points and minimize the perimeter.
• Second derivative test: To confirm that the critical point corresponds to a minimum.

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Summary of Steps

1. Express $y$ in terms of $x$ using the area equation: $y = \frac{36}{x}$.
2. Set up the perimeter function: $p = 2 \left( x + \frac{36}{x} \right)$.
3. Differentiate the perimeter function: $\frac{dp}{dx} = 2 \left( \frac{x^2 - 36}{x^2} \right)$.
4. Set $\frac{dp}{dx} = 0$ and solve for $x = 6$.
5. Confirm the minimum using the second derivative test.
6. Find the dimensions of the rectangle: $x = 6$ and $y = 6$.