Standard +0.3 This is a standard optimization problem requiring students to express area in terms of one variable using the circle equation, differentiate, and find the maximum. While it requires multiple steps (setting up the constraint, expressing area, differentiating, solving), the approach is routine and commonly practiced in A-level courses. The symmetry and use of Pythagoras/circle equation are straightforward, making this slightly easier than average.
13 A company is designing a logo. The logo is a circle of radius 4 inches with an inscribed rectangle. The rectangle must be as large as possible.
The company models the logo on an \(x - y\) plane as shown in the diagram.
\includegraphics[max width=\textwidth, alt={}, center]{85b10472-8149-4387-999f-2ef153f1a105-20_492_492_511_776}
Use calculus to find the maximum area of the rectangle.
Fully justify your answer.
Differentiates their expression for area. Condone one error
\(\frac{dA}{dx} = \frac{64-8x^2}{\sqrt{16-x^2}}\)
For maximum point \(\frac{dA}{dx} = 0\)
E1 (AO2.4)
Explains that their derivative equals zero for a maximum or stationary point
\(\frac{64-8x^2}{\sqrt{16-x^2}} = 0\)
A1 (AO1.1b)
Equates area derivative to zero and obtains correct value for either variable. CAO
\(x = 2\sqrt{2}\)
When \(x=2.8\), \(\frac{dA}{dx} = 0.448\); When \(x=2.9\), \(\frac{dA}{dx} = -1.191\); Therefore maximum
M1 (AO1.1a)
Completes a gradient test or uses second derivative of their area function to determine nature of stationary point
The maximum area is 32 sq in; maximum at \(x = 2\sqrt{2}\) or \(\theta = \frac{\pi}{4}\)
R1 (AO2.2a)
Deduces that the area is a maximum. Values need not be exact
Maximum area AWRT 32
B1 (AO3.2a)
Obtains maximum area with correct units
## Question 13:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Width of rectangle $= 2x$; Length of rectangle $= 2y$ | B1 (AO3.1b) | Identifies and clearly defines consistent variables for length and width. Can be shown on diagram |
| $A = 4xy$ | M1 (AO3.3) | Models the area of rectangle with an expression of the correct dimensions |
| $x^2 + y^2 = 16$ | M1 (AO1.1a) | Eliminates either variable to form a model for the area in one variable |
| $A = 4x\sqrt{16-x^2}$ | A1 (AO1.1b) | Obtains a correct equation to model the area in one variable |
| $\frac{dA}{dx} = 4\sqrt{16-x^2} - \frac{4x^2}{\sqrt{16-x^2}}$ | M1 (AO3.4) | Differentiates their expression for area. Condone one error |
| $\frac{dA}{dx} = \frac{64-8x^2}{\sqrt{16-x^2}}$ | | |
| For maximum point $\frac{dA}{dx} = 0$ | E1 (AO2.4) | Explains that their derivative equals zero for a maximum or stationary point |
| $\frac{64-8x^2}{\sqrt{16-x^2}} = 0$ | A1 (AO1.1b) | Equates area derivative to zero and obtains correct value for either variable. CAO |
| $x = 2\sqrt{2}$ | | |
| When $x=2.8$, $\frac{dA}{dx} = 0.448$; When $x=2.9$, $\frac{dA}{dx} = -1.191$; Therefore maximum | M1 (AO1.1a) | Completes a gradient test or uses second derivative of their area function to determine nature of stationary point |
| The maximum area is 32 sq in; maximum at $x = 2\sqrt{2}$ or $\theta = \frac{\pi}{4}$ | R1 (AO2.2a) | Deduces that the area is a maximum. Values need not be exact |
| Maximum area AWRT 32 | B1 (AO3.2a) | Obtains maximum area with correct units |
13 A company is designing a logo. The logo is a circle of radius 4 inches with an inscribed rectangle. The rectangle must be as large as possible.
The company models the logo on an $x - y$ plane as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{85b10472-8149-4387-999f-2ef153f1a105-20_492_492_511_776}
Use calculus to find the maximum area of the rectangle.\\
Fully justify your answer.\\
\hfill \mbox{\textit{AQA Paper 1 2018 Q13 [10]}}