Answer: \(x=225\) m and the minimum time is \(\dfrac{1600}{3}\) s, approximately \(533\) s.

(a) In right-angled triangle \(ABC\),

\(AC=\sqrt{300^2+x^2}.\)

The time for Joseph to walk from \(A\) to \(C\) is therefore

\(\frac{\sqrt{300^2+x^2}}{0.9}.\)

Since \(AE=400\) m and \(BC=x\) m,

\(CD=400-x.\)

The time for Joseph to walk from \(C\) to \(D\) is

\(\frac{400-x}{1.5}.\)

So the total time is

\(T=\frac{\sqrt{300^2+x^2}}{0.9}+\frac{400-x}{1.5}.\)

(b) Differentiate \(T\) with respect to \(x\):

\(\frac{dT}{dx} =\frac{1}{0.9}\cdot\frac{x}{\sqrt{300^2+x^2}}-\frac{1}{1.5}.\)

For a minimum, set \(\dfrac{dT}{dx}=0\):

\(\frac{x}{0.9\sqrt{300^2+x^2}}=\frac{1}{1.5}.\)

So

\(1.5x=0.9\sqrt{300^2+x^2}.\)

Square both sides:

\(2.25x^2=0.81(300^2+x^2).\)

Multiplying by \(100\),

\(225x^2=81(300^2+x^2).\)

Divide by \(9\):

\(25x^2=9(300^2+x^2).\)

Hence

\(16x^2=9\cdot300^2.\)

So

\(x=\frac{3\cdot300}{4}=225.\)

Now find \(T\) at \(x=225\):

\(AC=\sqrt{300^2+225^2}=\sqrt{140625}=375.\)

Therefore

\(T=\frac{375}{0.9}+\frac{175}{1.5} =\frac{1250}{3}+\frac{350}{3} =\frac{1600}{3}.\)

The minimum possible time is

\(\frac{1600}{3}\text{ s}\approx533\text{ s}.\)