Equation

Cost = $$c = c_1 x + c_2y$$ where $c_1$ is cost of travelling on water and $c_2$ is cost of travelling on land. As per the given condition, $$c_1 = \frac{c_2}{2}$$ Hence the equation becomes: $$c = \frac{cx}{2} + cy$$
$$=> c = c(x/2 + y)$$
Also, as per the given figure, the relation between $x$ and $y$ can be represented as follows: $$y^2 = (a-x)^2 + d^2$$
$$=> y = \sqrt{(a-x)^2 + d^2}                               \cdots 1$$                                                   
The objective is to minimize the cost function c. So $$\frac{d}{dx}c[\frac{x}{2}+ \sqrt{(a-x)^2 + d^2}] = 0$$ Solving the above differential equation: $$\frac{1}{2} - \frac{2(a - x)}{2\sqrt{(a-x)^2 + d^2}} = 0$$ $$=> \frac{1}{2} = \frac{(a - x)}{\sqrt{(a-x)^2 + d^2}}$$ Squaring both sides, we get: $$=> \frac{1}{4} = \frac{(a - x)^2}{(a-x)^2 + d^2}$$ $$=> 4(a-x)^2 = (a-x)^2 + d^2$$ $$=> 3(a-x)^2 = d^2$$ $$=> x = a - \frac{d}{\sqrt{3}}$$ From equation 1 above, $$y = \frac{2}{\sqrt3}d$$