Question
A company is designing a new carpet. The intended design of the carpet is in the shape of a rectangle with a semi-circle at each end.
The width of the rectangle is \( y \) metres, and the diameter of each semi-circle is \( x \) metres, with \( x > 0 \) and \( y \geq 0 \).
The company has decided that the perimeter of the carpet will be 20 metres and would like to maximize its area.
(a) Find an expression for the perimeter in terms of \( x \) and \( y \).
(b) Show that the area, \( A \) (m²), of the carpet can be expressed as:
\[ A = 10x – \frac{\pi x^2}{4}. \]
(c) Find \( \frac{dA}{dx} \).
(d) Hence, find the exact value of \( x \) for which the area is a maximum.
▶️ Answer/Explanation
(a) Finding an Expression for Perimeter
- The perimeter consists of two semicircles and two rectangle sides.
- The circumference of a full circle is \( \pi x \), so the total perimeter of two semicircles is \( \pi x \).
- The two straight sides contribute \( 2y \).
\[ \pi x + 2y = 20 \]
(b) Expressing the Area in Terms of \( x \)
- The area of the rectangle is \( A_{\text{rect}} = xy \).
- The area of the two semicircles (which form a full circle) is:
\[ A_{\text{circle}} = \frac{\pi x^2}{4}. \]
Thus, the total area:
\[ A = xy + \frac{\pi x^2}{4}. \]
Using the perimeter equation to express \( y \) in terms of \( x \):
\[ y = \frac{20 – \pi x}{2}. \]
Substituting \( y \) into the area equation:
\[ A = x \left( \frac{20 – \pi x}{2} \right) + \frac{\pi x^2}{4} \]
\[ = \frac{20x – \pi x^2}{2} + \frac{\pi x^2}{4} \]
Rewriting:
\[ = 10x – \frac{\pi x^2}{4}. \]
✔️ This matches the required expression.
(c) Finding \( \frac{dA}{dx} \)
Using differentiation:
\[ \frac{dA}{dx} = \frac{d}{dx} \left( 10x – \frac{\pi x^2}{4} \right) \]
\[ = 10 – \frac{\pi x}{2}. \]
(d) Finding the Maximum Area
To maximize the area, set \( \frac{dA}{dx} = 0 \):
\[ 10 – \frac{\pi x}{2} = 0. \]
Solving for \( x \):
\[ 10 = \frac{\pi x}{2} \]
\[ x = \frac{20}{\pi}. \]
Final Answer:
- The exact value of \( x \) that maximizes the area is \( \frac{20}{\pi} \).
…………………………..Markscheme…………………………..
- Correct perimeter equation: \( \pi x + 2y = 20 \).
- Correct area expression derivation: \( A = 10x – \frac{\pi x^2}{4} \).
- Correct derivative: \( \frac{dA}{dx} = 10 – \frac{\pi x}{2} \).
- Final answer: \( x = \frac{20}{\pi} \).
Question
The gradient of the normal to the curve:
\[ y = ax^2 + bx – 10 \]
at the point:
\[ T(2,4) \]
is:
\[ -\frac{1}{3} \]
Calculate the value of \( a \) and the value of \( b \).
▶️ Answer/Explanation
Detailed Solution
Step 1: Compute the Gradient of the Tangent
The derivative of the given function is:
\[ \frac{dy}{dx} = 2ax + b \]
At \( x = 2 \), the gradient of the normal is given as \( -\frac{1}{3} \).
The relationship between the tangent and normal gradients is:
\[ m_{\text{tangent}} \times m_{\text{normal}} = -1 \]
Substituting \( m_{\text{normal}} = -\frac{1}{3} \):
\[ m_{\text{tangent}} \times \left(-\frac{1}{3}\right) = -1 \]
\[ m_{\text{tangent}} = 3 \]
Since the derivative represents the gradient of the tangent:
\[ 3 = 2a(2) + b \]
\[ 3 = 4a + b \]
Step 2: Substitute the Point \( (2,4) \) into the Original Equation
Substituting \( x = 2 \) and \( y = 4 \) into:
\[ y = ax^2 + bx – 10 \]
\[ 4 = a(2)^2 + b(2) – 10 \]
\[ 4 = 4a + 2b – 10 \]
\[ 4a + 2b = 14 \]
\[ 2a + b = 7 \]
Step 3: Solve the System of Equations
We have the system:
\[ 4a + b = 3 \]
\[ 2a + b = 7 \]
Subtract the second equation from the first:
\[ (4a + b) – (2a + b) = 3 – 7 \]
\[ 2a = -4 \]
\[ a = -2 \]
Substituting \( a = -2 \) into \( 2a + b = 7 \):
\[ 2(-2) + b = 7 \]
\[ -4 + b = 7 \]
\[ b = 11 \]
Step 4: Final Answer
- Value of \( a \): \( -2 \)
- Value of \( b \): \( 11 \)
…………………………..Markscheme…………………………..
- Correct differentiation to find the gradient function.
- Correct substitution and solving of simultaneous equations.
- Final correct values: \( a = -2 \), \( b = 11 \).