To find the maximum product of the two lengths, we want to maximize both lengths individually. Let's call the length of the cable $L$.

We want to cut the cable into two parts, $x$ and $L-x$, where $x$ represents the length of one part and $L-x$ represents the length of the other part.

To find the maximum product, we want to maximize $x(L-x)$. We can apply the AM-GM inequality, which states that for any positive real numbers $a$ and $b$, the arithmetic mean is greater than or equal to the geometric mean: $\frac{a+b}{2} \geq \sqrt{ab}$.

In this case, let $a = x$ and $b = L-x$. Thus, $\frac{x+(L-x)}{2} \geq \sqrt{x(L-x)}$.

Simplifying, we have $\frac{L}{2} \geq \sqrt{x(L-x)}$.

To find the maximum value of $\sqrt{x(L-x)}$, we need to find the maximum value of $x(L-x)$.

Since the cable length $L$ is fixed, the product $x(L-x)$ is a quadratic function with respect to $x$.

The maximum value of a quadratic function occurs at the vertex, which is the point $(\frac{-b}{2a}, f(\frac{-b}{2a}))$.

In this case, $a = -1$, $b = L$, and $c = 0$. So the vertex is $(\frac{-L}{2(-1)}, f(\frac{-L}{2(-1)}))$, which simplifies to $(\frac{L}{2}, f(\frac{L}{2}))$.

Thus, the maximum value of $x(L-x)$ occurs when $x = \frac{L}{2}$.

Therefore, the maximum product of the two lengths is $\sqrt{x(L-x)} = \sqrt{\frac{L}{2} \cdot \frac{L}{2}} = \frac{L}{2} \cdot \frac{L}{2} = \frac{L^2}{4}$.

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