Find the number that will maximise the intake.

Let $N$ be the number of passengers.
The problem states that $50 \le N \le 70$.
If $N$ is exactly 50, the fare is \600,000$. If$N$exceeds 50, the fare is reduced. Let$x$be the excess number of passengers, so$x = N - 50$. The reduction in fare is$$10,000 \times x$. The fare per passenger,$F(N)$, is:

$$F(N) = 600,000 - 10,000(N - 50)$$

The total intake (revenue), $I(N)$, is the number of passengers multiplied by the fare per passenger:

$$I(N) = N \times F(N)$$

Substitute the expression for $F(N)$:

$$I(N) = N [600,000 - 10,000(N - 50)]$$

Expand and simplify the expression:

$$I(N) = N [600,000 - 10,000N + 500,000]$$ $$I(N) = N [1,100,000 - 10,000N]$$ $$I(N) = 1,100,000N - 10,000N^2$$

To find the number of passengers that maximizes the intake, we need to find the derivative of $I(N)$ and set it to zero:

$$\frac{dI}{dN} = \frac{d}{dN}(1,100,000N - 10,000N^2)$$ $$\frac{dI}{dN} = 1,100,000 - 20,000N$$

$$1,100,000 - 20,000N = 0$$ $$20,000N = 1,100,000$$ $$N = \frac{1,100,000}{20,000}$$ $$N = 55$$

The value $N=55$ is within the allowed range of $50 \le N \le 70$.
To confirm it's a maximum, we can use the second derivative test:

$$\frac{d^2I}{dN^2} = \frac{d}{dN}(1,100,000 - 20,000N)$$ $$\frac{d^2I}{dN^2} = -20,000$$

Since the second derivative is negative ($-20,000 < 0$), $N=55$ corresponds to a maximum intake.

The number of passengers that will maximize the intake is 55.

$$55 \text{ passengers}$$