Sample Answer
Optimisation and Application of Calculus in Geometrical Problems
Mathematical optimisation is a powerful tool used to solve real-world geometry problems where a fixed quantity must be divided or allocated efficiently. The following discussion explores three distinct problems that rely on logical reasoning, algebra, and calculus to find accurate and meaningful results.
Question 1: Combined Areas of an Equilateral Triangle and a Circle
A wire measuring 100 cm is divided into two parts. One part is bent to form an equilateral triangle, while the other is shaped into a circle. The task is to determine the conditions under which the total combined area of both shapes is at its maximum and minimum.
Let the portion used for the triangle be x cm, leaving 100 − x cm for the circle. The side length of the triangle will be x/3, giving it an area of (√3/36) × x². For the circle, the radius is (100 − x) / (2π), giving an area of (100 − x)² / (4π). The total area therefore equals (√3/36) × x² + (100 − x)² / (4π).
Differentiating this expression and setting the result equal to zero identifies the critical point where x ≈ 62.3 cm. This means 62.3 cm of the wire should form the triangle and 37.7 cm should form the circle. Testing this point confirms it provides the minimum total area. The maximum area occurs when the entire wire is used for one shape only, with the circle producing the larger area since, for a given perimeter, the circle always encloses the maximum possible area.
Question 2: Using Euler’s Method to Approximate a Differential Equation
The differential equation dx/dt = t² + x × e^(-t), with an initial condition x(0.1) = −1, can be solved approximately using Euler’s numerical method. This approach estimates values of x at chosen time steps, such as t = 0.5 and t = 2.5, using a spreadsheet or calculator.
Euler’s formula is x_(n+1) = x_n + h × f(t_n, x_n), where h is the step size. Repeated substitution provides an estimated trajectory of x as t increases. Using small time increments (for example, h = 0.1) allows the spreadsheet to calculate x values with acceptable accuracy. Results generally show that as t increases, the exponential term e^(-t) decreases, reducing its influence and allowing t² to dominate the rate of change of x. This demonstrates how numerical methods like Euler’s are useful for understanding complex systems when an exact analytical solution is impractical.
Question 3: Ladder Leaning Against a Wall and a Cube
A ladder 3.75 metres long leans against a vertical wall and just touches the outer top edge of a cubic packing case that rests against the wall. The problem is to determine how far up the wall the top of the ladder reaches.
If the cube has a side length of one metre, a right-angled triangle can be visualised between the wall, the ground, and the ladder. The ladder passes over the cube’s top corner, forming two smaller right triangles. Using geometric reasoning or coordinate geometry, it can be shown that the top of the ladder reaches approximately 3.4 metres up the wall. The exact height depends on the cube’s size, but the key principle lies in applying Pythagoras’ theorem and proportional reasoning to determine the correct vertical height.
