Let $$n \geq 2$$ be a natural number and $$f : [0, 1] \to \mathbb{R}$$ be the function defined by

$$f(x) = \begin{cases} n(1 - 2nx) & \text{if } 0 \leq x \leq \frac{1}{2n} \\ 2n(2nx - 1) & \text{if } \frac{1}{2n} \leq x \leq \frac{3}{4n} \\ 4n(1 - nx) & \text{if } \frac{3}{4n} \leq x \leq \frac{1}{n} \\ \frac{n}{n-1}(nx - 1) & \text{if } \frac{1}{n} \leq x \leq 1 \end{cases}$$

If n is such that the area of the region bounded by the curves $$x = 0$$, $$x = 1$$, $$y = 0$$ and $$y = f(x)$$ is 4, then the maximum value of the function f is

The function $$f:[0,1]\rightarrow \mathbb{R}$$ is piece-wise linear and reaches its highest ordinate at the end-points of each linear “saw-tooth”.
At every such peak the value is $$n$$, so the required maximum of $$f$$ will simply be $$n$$ once we know the admissible value of $$n$$.

To determine $$n$$ we use the information that the area enclosed by the axes, the line $$x=1$$ and the curve $$y=f(x)$$ equals $$4$$.
That area is the definite integral $$A(n)=\displaystyle\int_{0}^{1} f(x)\,dx$$. Evaluate it interval by interval.

Interval 1:  $$0\le x\le\dfrac{1}{2n}$$
$$f(x)=n(1-2nx)=n-2n^{2}x$$

$$I_{1}=\int_{0}^{1/(2n)} (n-2n^{2}x)\,dx =\left[nx-n^{2}x^{2}\right]_{0}^{1/(2n)} =\dfrac{1}{2}-\dfrac14=\dfrac14$$

Interval 2:  $$\dfrac{1}{2n}\le x\le\dfrac{3}{4n}$$
$$f(x)=2n(2nx-1)=4n^{2}x-2n$$

$$I_{2}=\int_{1/(2n)}^{3/(4n)} (4n^{2}x-2n)\,dx =\left[2n^{2}x^{2}-2nx\right]_{1/(2n)}^{3/(4n)} =\dfrac18$$

Interval 3:  $$\dfrac{3}{4n}\le x\le\dfrac{1}{n}$$
$$f(x)=4n(1-nx)=4n-4n^{2}x$$

$$I_{3}=\int_{3/(4n)}^{1/n} (4n-4n^{2}x)\,dx =\left[4nx-2n^{2}x^{2}\right]_{3/(4n)}^{1/n} =\dfrac18$$

Interval 4:  $$\dfrac{1}{n}\le x\le1$$
$$f(x)=\dfrac{n}{\,n-1\,}(nx-1) =\dfrac{n^{2}}{\,n-1\,}x-\dfrac{n}{\,n-1\,}$$

$$I_{4}=\int_{1/n}^{1} \left(\dfrac{n^{2}}{\,n-1\,}x-\dfrac{n}{\,n-1\,}\right)\!dx =\left[\dfrac{n^{2}x^{2}}{2(n-1)}-\dfrac{nx}{\,n-1\,}\right]_{1/n}^{1} =\dfrac{n(n-2)+1}{2(n-1)}$$

Total area:
$$A(n)=I_{1}+I_{2}+I_{3}+I_{4} =\dfrac14+\dfrac18+\dfrac18+\dfrac{n(n-2)+1}{2(n-1)} =\dfrac12+\dfrac{n(n-2)+1}{2(n-1)}$$

The problem states $$A(n)=4$$, so

$$\dfrac12+\dfrac{n(n-2)+1}{2(n-1)}=4 \quad\Longrightarrow\quad 1+\dfrac{n(n-2)+1}{n-1}=8$$

$$n(n-2)+1=7(n-1) \;\;\Longrightarrow\;\; n^{2}-9n+8=0 \;\;\Longrightarrow\;\; n=\dfrac{9\pm\sqrt{49}}{2}=8\ \text{or}\ 1$$

Given $$n\ge2$$, we must take $$n=8$$.

Because the highest ordinate of $$f$$ is always $$n$$, the maximum value of the function under these conditions is
$$\boxed{8}$$