2.10 Q-5

Question Statement

Find the dimensions of a rectangle with the largest area, given that its perimeter is 120 cm.

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Background and Explanation

This problem involves finding the dimensions of a rectangle that maximize its area, given a fixed perimeter. The perimeter constraint provides a relationship between the length and breadth of the rectangle. By expressing the area in terms of one variable and differentiating it, we can find the dimensions that maximize the area.

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Solution

1. Define the variables:

   Let $x$ be the length of the rectangle (in cm) and $y$ be the breadth. The perimeter of the rectangle is given as 120 cm, so the relationship between $x$ and $y$ is:

$$2x + 2y = 120$$

Dividing through by 2:

$$x + y = 60$$

Hence, $y = 60 - x$.

2. Set up the area function:

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

$$A = x \cdot y = x \cdot (60 - x)$$

This is the function we need to maximize.

3. Differentiate the area function:

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

$$\frac{dA}{dx} = \frac{d}{dx}[x(60 - x)]$$

Applying the product rule:

$$\frac{dA}{dx} = (60 - x) + x(-1) = 60 - 2x$$

4. Set the derivative equal to zero:

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

$$60 - 2x = 0$$

Solving for $x$:

$$2x = 60 \quad \Rightarrow \quad x = 30$$

5. Check the second derivative:

   To confirm that $x = 30$ gives a maximum, compute the second derivative:

$$\frac{d^2A}{dx^2} = \frac{d}{dx}(60 - 2x) = -2$$

Since $\frac{d^2A}{dx^2} = -2$, which is negative, $x = 30$ gives a maximum.

6. Find the dimensions of the rectangle:

   The breadth of the rectangle is:

$$y = 60 - 30 = 30$$

Hence, the dimensions of the rectangle that maximize the area are 30 cm by 30 cm.

Thus, the rectangle with the largest area has side lengths of 30 cm by 30 cm.

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Key Formulas or Methods Used

• Perimeter of a rectangle: $2x + 2y = 120$
• Area of a 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 $y$ in terms of $x$ using the perimeter equation: $y = 60 - x$.
2. Set up the area function: $A = x \cdot (60 - x)$.
3. Differentiate the area function with respect to $x$ to get $\frac{dA}{dx} = 60 - 2x$.
4. Set $\frac{dA}{dx} = 0$ and solve for $x = 30$.
5. Confirm the maximum using the second derivative test.
6. Find the dimensions: $x = 30$ and $y = 30$, so the rectangle is 30 cm by 30 cm.

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