2.10 Q-8

Question Statement

Find the dimensions of a rectangular garden with a perimeter of 80 meters, such that its area is maximized.

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Background and Explanation

In this problem, we are asked to maximize the area of a rectangular garden given a fixed perimeter. We use the relationship between the perimeter and the sides of the rectangle to express one dimension in terms of the other. Then, we apply calculus to maximize the area by finding the critical points and using the second derivative to confirm that the critical point corresponds to a maximum.

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Solution

1. Define the variables:

   Let $x$ be the length and $y$ be the breadth of the rectangular garden. The perimeter $p$ of the rectangle is given by:

$$p = 2x + 2y = 80$$

Simplifying:

$$x + y = 40 \quad \text{(Equation 1)}$$

This equation relates the length and breadth.

2. Express the area:

   The area $A$ of the rectangle is given by:

$$A = x \cdot y$$

Using Equation 1, we can express $y$ as:

$$y = 40 - x$$

Substituting into the area formula:

$$A = x \cdot (40 - x) = 40x - x^2$$

This is the area function in terms of $x$.

3. Differentiate the area function:

   To find the value of $x$ that maximizes the area, we differentiate the area function with respect to $x$:

$$\frac{dA}{dx} = 40 - 2x$$

Setting the derivative equal to zero to find the critical points:

$$40 - 2x = 0 \quad \Rightarrow \quad x = 20$$

4. Check the second derivative:

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

$$\frac{d^2A}{dx^2} = -2$$

Since the second derivative is negative, $x = 20$ corresponds to a maximum.

5. Find the breadth $y$:

   Now, substitute $x = 20$ into Equation 1 to find the breadth:

$$y = 40 - 20 = 20$$

Thus, the dimensions of the garden that will maximize the area are 20 meters by 20 meters.

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Key Formulas or Methods Used

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

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Summary of Steps

1. Express the length and breadth in terms of each other using the perimeter equation: $x + y = 40$.
2. Write the area function as $A = x(40 - x)$.
3. Differentiate the area function: $\frac{dA}{dx} = 40 - 2x$.
4. Set $\frac{dA}{dx} = 0$ and solve for $x = 20$.
5. Confirm the maximum using the second derivative: $\frac{d^2A}{dx^2} = -2$.
6. Find the breadth: $y = 40 - 20 = 20$.
7. The dimensions that maximize the area are 20 meters by 20 meters.

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