Maximum Value of Piecewise Function Bounding Given Area

To find the maximum value of the function $f(x)$, we first compute the total area of the region bounded by $x = 0$, $x = 1$, $y = 0$, and $y = f(x)$.

The function $f(x)$ is non-negative on $[0,1]$ and is defined piecewise. We can calculate the area $A$ bounded by $y = f(x)$ and the $x$-axis by dividing the interval $[0,1]$ into four sub-intervals:

1. For $x \in \left[0, \frac{1}{2n}\right]$:
   $I_1 = \int_0^{\frac{1}{2n}} n(1-2nx) \, dx = n \left[ x - nx^2 \right]_0^{\frac{1}{2n}} = n \left( \frac{1}{2n} - \frac{n}{4n^2} \right) = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$

2. For $x \in \left[\frac{1}{2n}, \frac{3}{4n}\right]$:
   $I_2 = \int_{\frac{1}{2n}}^{\frac{3}{4n}} 2n(2nx-1) \, dx = 2n \left[ nx^2 - x \right]_{\frac{1}{2n}}^{\frac{3}{4n}} = 2n \left( \left( \frac{9}{16n} - \frac{3}{4n} \right) - \left( \frac{1}{4n} - \frac{1}{2n} \right) \right) = 2n \left( -\frac{3}{16n} + \frac{1}{4n} \right) = \frac{1}{8}$

3. For $x \in \left[\frac{3}{4n}, \frac{1}{n}\right]$:
   $I_3 = \int_{\frac{3}{4n}}^{\frac{1}{n}} 4n(1-nx) \, dx = 4n \left[ x - \frac{nx^2}{2} \right]_{\frac{3}{4n}}^{\frac{1}{n}} = 4n \left( \left( \frac{1}{n} - \frac{1}{2n} \right) - \left( \frac{3}{4n} - \frac{9}{32n} \right) \right) = 4n \left( \frac{1}{2n} - \frac{15}{32n} \right) = \frac{1}{8}$

4. For $x \in \left[\frac{1}{n}, 1\right]$:
   $I_4 = \int_{\frac{1}{n}}^{1} \frac{n}{n-1}(nx-1) \, dx = \frac{n}{n-1} \left[ \frac{nx^2}{2} - x \right]_{\frac{1}{n}}^{1} = \frac{n}{n-1} \left( \left( \frac{n}{2} - 1 \right) - \left( \frac{1}{2n} - \frac{1}{n} \right) \right) = \frac{n}{n-1} \cdot \frac{(n-1)^2}{2n} = \frac{n-1}{2}$

Summing up the individual areas gives the total area $A$:
$A = I_1 + I_2 + I_3 + I_4 = \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{n-1}{2} = \frac{1}{2} + \frac{n-1}{2} = \frac{n}{2}$

We are given that the total area $A = 4$: $\frac{n}{2} = 4 \implies n = 8$

Since $f(x)$ is piecewise linear on each sub-interval, its maximum value must occur at one of the boundary points:

• $f(0) = n = 8$
• $f\left(\frac{1}{2n}\right) = 0$
• $f\left(\frac{3}{4n}\right) = n = 8$
• $f\left(\frac{1}{n}\right) = 0$
• $f(1) = n = 8$

Thus, the maximum value of the function $f$ is $n = 8$.