Çözüldü Differentiation - Optimization - Quadratic Equations

From The West Shore School District:

A power line is needed to connect a power station on the shore of a river to an island 4 miles downstream and 1 mile offshore. Find the minimum cost for such a line given that is costs $50,000 per mile to lay wire under the water and $30,000 per mile to lay wire underground.
https://www.wssd.k12.pa.us/downloads/calculus optimization problems solutions.pdf
(The last question without solution.)


https://i72.servimg.com/u/f72/19/97/10/39/powerl10.png

x: The length of the underground cable
y: The length of the cable under the river
y(x) = [ 1^2 + (4 - x)^2 ]^0.5 = [ 1 + (4 - x)^2 ]^0.5....(I)
L(x): Total length of the cable = x + y(x)
L(x) = x + [ 1^2 + (4 - x)^2 ]^0.5
C(x): Total cost of the cable
C(x) = 30.000·x + 50.000·y(x)
C '(x) = 30.000 + 50.000·y '(x) = 0 ⇒ y '(x) = -3 / 5....(II)
Differentiating (I); y '(x) = -(4 - x) / { [ 1 + (4 - x)^2 ]^0.5 }....(III)
From the equations (II) and (III), -3 / 5 = -(4 - x) / { [ 1 + (4 - x)^2 ]^0.5 }
4 - x = p....(IV)
3 / 5 = p / √(1 + p^2)
9 / 25 = p^2 / (1 + p^2)
p = ∓3 / 4....(V)
Putting (V) values in (IV) and then in (I),
p1 = -3 / 4 ⇒ x = 4 - (-3 / 4) = 19 / 4 miles ⇒ y1(19 / 4) = 5 / 4
p2 = 3 / 4 ⇒ x2 = 4 - 3 / 4 = 13 / 4 miles ⇒ y2(13 / 4) = 5 / 4
Minimum Cost: 30.000·(13 / 4) + 50.000·(5 / 4) = $160.000.