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, $$y = \frac{2}{\sqrt3}d$$
$$ => c = c(x/2 + y)$$
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, $$y = \frac{2}{\sqrt3}d$$
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