Exercise 4 · Option A · Surface optimisation
Minimise the sum of the areas of a circle and a square when the total length of the two railings is fixed.
Maximum score · 2.5 pointsIn the exam you must choose one of the two options (A or B). Here you have Option A worked out; Option B is on the following page.
- The mayor of a town in Catalonia asks the municipal architect to design a children's playground. There must be two delimited spaces: one for a water tap (circular in shape) and another for a tool shed (square in shape). Each space is delimited with a wrought-iron railing. Knowing that the two railings measure exactly 10 m in total length, what should the length of the railing of each space be so that the sum of the areas of the two spaces is as small as possible? What is this minimum area? 2.5 p
Step-by-step solution
Key idea: each railing is the perimeter of its figure. We express the two areas as a function of a single variable (the length of the circular railing), using the constraint that they add up to 10 m, and we minimise with the derivative.
· Setting up with one variable
Let $c$ be the length of the circular railing (in metres); then the square railing is $10-c$. Each railing is the perimeter of the figure:
The function to minimise, with $c\in[0,10]$, is:
· Derivative and critical point
We differentiate and set it to zero:
Since $A''(c)=\dfrac{1}{2\pi}+\dfrac{1}{8}>0$, the function is convex and this critical point is a minimum. The square railing measures:
· Minimum area
We substitute into $A(c)$. It is convenient to simplify each term: