FPM(2023) January Paper 01 que 8 and solution

Question 8

Figure 1 shows a badge, shown shaded, made from two identical rectangles, ABCD and DEFG, and a sector DCG of a circle with centre D.

Each rectangle measures $x\text{ cm}$ by $y\text{ cm}$.

The radius of the sector is $x\text{ cm}$ and the angle $\angle CDG$ is $0.5$ radians.

The area of the badge is $50\text{ cm}^2$

The perimeter of the badge is $P\text{ cm}$.

(a) Show that

$$ P = 2x + \frac{100}{x} $$

(5)

Given that $x$ can vary,

(b) use calculus, to find the exact value of $x$ for which $P$ is a minimum.

Justify that this value of $x$ gives a minimum value for $P$

(6)

(c) Find the minimum value of $P$

Give your answer in the form $k\sqrt{2}$, where $k$ is an integer to be found.

(2)

Solution

Derivation of Perimeter $P$

The badge consists of two identical rectangles $ABCD$ and $DEFG$, and a sector $DCG$ of a circle with center $D$.

• Each rectangle has dimensions $x\ \text{cm}$ by $y\ \text{cm}$.
• The radius of the sector is $x\ \text{cm}$.
• The angle of the sector $\angle CDG = 0.5\ \text{radians}$.
• The total area of the badge is given as $50\ \text{cm}^2$.
• The perimeter of the badge is $P\ \text{cm}$.

Step 1: Expression for Total Area

The area of the badge includes:

1. The areas of the two rectangles:

   $$ 2 \times \text{Area of one rectangle} = 2(x \cdot y) = 2xy $$

2. The area of the sector:

   $$ \text{Area of sector} = \frac{1}{2} \cdot x^2 \cdot 0.5 = 0.25x^2 $$

Thus, the total area is:

$$ \text{Total Area} = 2xy + 0.25x^2 $$

Given that the total area is $50\ \text{cm}^2$, we have:

$$ 2xy + 0.25x^2 = 50 $$

Step 2: Expression for Perimeter $P$

The perimeter of the badge includes:

1. The two vertical sides of the rectangles: $2y$
2. The two horizontal sides of the rectangles (excluding the part covered by the arc): $x + x = 2x$
3. The arc of the sector:

   $$ \text{Arc length} = x \cdot 0.5 = 0.5x $$

Thus, the total perimeter is:

$$ P = 2y + 2x + 0.5x = 2y + 2.5x $$

From the area equation, solve for $y$:

$$ y = \frac{50 - 0.25x^2}{2x} $$

Substitute $y$ into the expression for $P$:

$$ P = 2\left(\frac{50 - 0.25x^2}{2x}\right) + 2.5x $$

Simplify:

$$ P = \frac{50 - 0.25x^2}{x} + 2.5x $$

$$ P = \frac{50}{x} - \frac{0.25x^2}{x} + 2.5x $$

$$ P = \frac{50}{x} + 2x $$

Conclusion

The perimeter of the badge is:

$$ P = 2x + \frac{100}{x} $$

(b) Using Calculus to Minimize $P$

The expression for $P$ is:

$$ P = 2x + \frac{100}{x} $$

To find the value of $x$ that minimizes $P$, we calculate the derivative of $P$ with respect to $x$ and set it equal to zero:

$$ \frac{dP}{dx} = 2 - \frac{100}{x^2} $$

Set $\frac{dP}{dx} = 0$:

$$ 2 - \frac{100}{x^2} = 0 $$

$$ \frac{100}{x^2} = 2 $$

$$ x^2 = \frac{100}{2} = 50 $$

$$ x = \sqrt{50} = 5\sqrt{2} \quad (\text{since } x > 0) $$

Justification for Minimum

To confirm that $P$ has a minimum at $x = 5\sqrt{2}$, we compute the second derivative of $P$:

$$ \frac{d^2P}{dx^2} = \frac{200}{x^3} $$

Since $x > 0$, $\frac{d^2P}{dx^2} > 0$, indicating that $P$ has a local minimum at $x = 5\sqrt{2}$.

(c) Finding the Minimum Value of $P$

Substitute $x = 5\sqrt{2}$ into the expression for $P$:

$$ P = 2x + \frac{100}{x} $$

$$ P = 2(5\sqrt{2}) + \frac{100}{5\sqrt{2}} $$

$$ P = 10\sqrt{2} + \frac{100}{5\sqrt{2}} $$

$$ P = 10\sqrt{2} + \frac{20}{\sqrt{2}} $$

$$ P = 10\sqrt{2} + 10\sqrt{2} $$

$$ P = 20\sqrt{2} $$

Conclusion

The minimum value of $P$ is:

$$ P_{\text{min}} = 20\sqrt{2} $$

where $k = 20$.