A natural gas line runs east-west. A lown wants to connect two new housing developments to the line by running lines from a single point on the existing line to the two developments, One development, D1, is 3 km due south of the existing line; the other development, D2, is 4 km due south of the existing line and 5 km east of the first development. Find the place on the existing line to make the connection to minimize the total length of the new line.
My approach was the following:
Call the two distances from the center of the east-west pipeline $x$ and $5-x$, respectively. We get both hypotenuses, call them h1 and h2:
$$h1 = \sqrt{9+x^2}$$
$$h2 = \sqrt{16+(5-x)^2}$$
Then, we take the derivative of $ h1 + h2 $, and set it to zero to find the inflection point, x, and thus, the minimum possible total distance. The problem is that this derivative is too big for my smooth brain, and I'm completely stuck.
My teacher said that we should take some approach using related triangles. She made the assumption that the two triangles are related, and then we can easily solve the hypotenuses. But there was no justification. How can we make this assumption?
Best Answer
Approach 1
As mentioned in the comments
Given : AB= 3km , BE = 5km , EF= 4km and $\angle B,\angle E = 90$
Construction : Extend AB to C such that $AB = BC$ .
It is clear that $\triangle ABD\cong\triangle CBD$ $[BD = BD , BC = AB , \angle B = 90]$
By CPCT $AD = DC$
Therefore $ AD + DF = CD + DF $
The shortest possible distance between $C$ and $F$ is a line . Therefore , $D$ must lie on the line $CF$
Therefore, $\angle CDB = \angle BDA = \angle FDE$ [ By CPCT , Opposite angles]
With this we can prove similarity and reach a solution.
Approach 2
Continuing your approach taking the derivative of $h1+h2$
$$\biggl(\sqrt{9+x^2}+\sqrt{16+(5-x^2)}\biggr)'$$ $$\frac{x}{\sqrt{9+x^2}} + \frac{x-5}{\sqrt{16+(5-x^2)}}=0$$ $$\frac{x}{\sqrt{9+x^2}}= -\frac{x-5}{\sqrt{16+(5-x^2)}}$$ Cross multiplying $$x\sqrt{16+(5-x^2)} = -(x-5)\sqrt{9+x^2}$$ Squaring both sides $$x^2(41+x^2-10x) = (x^2 +25-10x)(9+x^2)$$ Expanding both sides and factorising leads us to the expression $$(7x-15)(x+15)=0$$ $$x=\frac{15}{7}$$