Hello Lavender V,
Let's let $0\le x\le 5$ represent the length of segment $\overline{DP}$ in meters. And so the total length of cable as a function of $x$ is then obtained via the Pythagorean theorem as:
$$L(x)=(5-x)+\sqrt{x^2+2^2}+\sqrt{x^2+3^2}$$
Now, we may differentiate with respect to $x$ and equate the result to zero to obtain the critical value(s):
$$L'(x)=-1+\frac{x}{\sqrt{x^2+4}}+\frac{x}{\sqrt{x^2+9}}=0$$
Now, since we are thankfully allowed to approximate, we may use Newton's method here to approximate the root of this equation.
Let:
$$f(x)=-1+\frac{x}{\sqrt{x^2+4}}+\frac{x}{\sqrt{x^2+9}}$$
We need to compute the first derivative, and so for the last two terms, consider:
$$g(x)=\frac{x}{\sqrt{x^2+k}}$$
Using the quotient, power and chain rules, we obtain:
$$g'(x)=\frac{\sqrt{x^2+k}(1)-x\dfrac{x}{\sqrt{x^2+k}}}{\left(\sqrt{x^2+k} \right)^2}=\frac{k}{\left(x^2+k \right)^{\frac{3}{2}}}$$
And so we may conclude that:
$$f'(x)=\frac{4}{\left(x^2+4 \right)^{\frac{3}{2}}}+\frac{9}{\left(x^2+9 \right)^{\frac{3}{2}}}$$
Newton's method gives us the recursion:
$$x_{n+1}=x_{n}-\frac{f\left(x_n \right)}{f'\left(x_n \right)}$$
And so using the definition of $f$ and its derivative, we have:
$$x_{n+1}=x_{n}-\frac{-1+\dfrac{x_n}{\sqrt{x_n^2+4}}+\dfrac{x_n}{\sqrt{x_n^2+9}}}{\dfrac{4}{\left(x_n^2+4 \right)^{\frac{3}{2}}}+\dfrac{9}{\left(x_n^2+9 \right)^{\frac{3}{2}}}}$$
Now, using a computer or CAS, and $x_0=1$, we find:
$$x_1\approx1.36825580018177$$
$$x_2\approx1.40658793851904$$
$$x_3\approx1.40699724822532$$
$$x_4\approx1.40699729457262$$
$$x_5\approx1.40699729457262$$
This is accurate to 15 digits, more than enough to get an approximation for the minimum of $L$ to two decimal places. Now, if we note that:
$$f'(x)=L''(x)>0$$
For all real values of $x$, then by the second derivative test, we may conclude that this critical value is at the global minimum, and we need not check the boundaries.
Thus, we may conclude:
$$L_{\min}\approx L(1.40699729457262)\approx9.35\text{ m}$$
