"A 1x1 metre tile costs £45.00. A garden owner has £900 (for tiles) and wants to build a new swimming pool with the base of the pool tiled, as well as the sides. The pool is one metre deep and is twice as long as it is wide. What is the maximum width/length that the pool can be?"
So essentially, we are looking for a rectangle where we tile the area inside the rectangle, as well as the perimeter. How would this question be set up given the constraints mentioned? Hope this is clear!
Firstly, let's say the pool is $x$ metres wide. It is given that it is twice as long, hence its length is $2x$.
This suffices to find the total area that must be covered as a function of $x$.
The sides of the pool are $4$ rectangles of width $1$. Two of them have length $x$, and the other $2$ have length $2x$. So,
$$\text{Sides Area}=1\cdot x+1\cdot x+1\cdot 2x+1\cdot 2x=6x \space [m^2]$$
The floor has length $2x$ and width $x$, meaning its area is:
$$\text{Floor Area}=2x\cdot x=2x^2 \space [m^2]$$
In total, the area which the tiles should cover is:
$$\text{Area}=\text{Sides Area}+\text{Floor Area}=2x^2+6x \space [m^2]$$
We know that the cost for a $1\times 1$ metre tile is $45$£, implying the cost as a function of $x$ would be:
$$\text{Cost}(x)=45(2x^2+6x)=90x^2+270x$$
The gardener only has $900$£, so we need this cost to be below that:
$$90x^2+270x\leq 900 \implies x^2+3x\leq 10 \implies x^2+3x-10 \leq 0 \implies (x+5)(x-2) \leq 0$$ $$\implies (x+5\leq 0 \space \text{and} \space x-2 \geq 0) \space \text{or} \space (x+5\geq 0 \space \text{and} \space x-2 \leq 0)$$ $$\implies x \geq -5 \space \text{and} \space x\leq 2 \implies -5\leq x\leq 2$$
$x$ represents length, so more strictly $0<x\leq 2$.
Thus, the maximum width is $2$, and the maximum length is $2\cdot 2=4$.