2.10 Q-10

Question Statement

Find the dimensions of the rectangle of maximum area that fits inside a semi-circle of radius 8 cm.

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Background and Explanation

This problem involves finding the dimensions of a rectangle that fits inside a semi-circle. The area of the rectangle is a function of its width and height, and the goal is to maximize this area. To do this, we will use calculus, specifically optimization techniques, to find the dimensions that result in the maximum area.

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Solution

1. Define the variables:
   • Let $x$ be the width of the rectangle.
   • The length of the rectangle is given as $2x$, since the rectangle is inscribed in the semi-circle.
   • The area $y$ of the rectangle is:

$$y = 4x^2(64 - x^2)$$

This equation accounts for the relationship between the width of the rectangle and the radius of the semi-circle.

2. Simplify the equation for the area: Expanding the expression for $y$:

$$y = 256x^2 - 4x^4$$

3. Differentiate the area function: To find the value of $x$ that maximizes the area, we first differentiate $y$ with respect to $x$:

$$\frac{dy}{dx} = 512x - 16x^3$$

Set this equal to zero to find the critical points:

$$512x - 16x^3 = 0$$

Factor the equation:

$$16x(32 - x^2) = 0$$

This gives us two solutions:

• $x = 0$ (which is not meaningful since we want a non-zero rectangle),
• $x^2 = 32$ or $x = 4\sqrt{2}$.

4. Second derivative test: To confirm that this critical point corresponds to a maximum, we calculate the second derivative of $y$:

$$\frac{d^2y}{dx^2} = 512 - 48x^2$$

Substituting $x = 4\sqrt{2}$:

$$\frac{d^2y}{dx^2} = 512 - 48(32) = -1024$$

Since the second derivative is negative, this confirms that $x = 4\sqrt{2}$ corresponds to a maximum.

5. Calculate the dimensions of the rectangle:
   • The width of the rectangle is $x = 4\sqrt{2}$ cm.
   • The length of the rectangle is $2x = 8\sqrt{2}$ cm.

Thus, the dimensions of the rectangle that maximize the area are:

• Width: $4\sqrt{2}$ cm
• Length: $8\sqrt{2}$ cm

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Key Formulas or Methods Used

• Area of the rectangle: $y = 4x^2(64 - x^2)$
• First derivative: To find the critical points for maximum area.
• Second derivative test: To confirm whether the critical point is a maximum or minimum.

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Summary of Steps

1. Express the area $y = 256x^2 - 4x^4$.
2. Differentiate to find $\frac{dy}{dx} = 512x - 16x^3$.
3. Set $\frac{dy}{dx} = 0$ and solve for $x = 4\sqrt{2}$.
4. Use the second derivative $\frac{d^2y}{dx^2}$ to confirm a maximum at $x = 4\sqrt{2}$.
5. Calculate the width $x = 4\sqrt{2}$ cm and length $2x = 8\sqrt{2}$ cm.

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