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Min max problem: length of pipe from 2 towns to the river

  • Thread starter Karol
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  • #1
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Homework Statement


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Homework Equations


Minimum/Maximum occurs when the first derivative=0

The Attempt at a Solution


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$$l_1^2+l_2^2=(a^2+x^2)+[b^2+(d-x)^2]=a^2+b^2+x^2+\left[ \sqrt{c^2-(b-a)^2}-x \right]^2$$
$$(l_1^2+l_2^2)'=2x+2\left[ \sqrt{c^2-(b-a)^2}-x \right](-1)$$
$$(l_1^2+l_2^2)'=0~~\rightarrow~~x=\frac{1}{2}\sqrt{c^2-(b-a)^2}=\frac{d}{2}$$
The minimum (or maximum?) distance is when the station is in the middle of the vertical distance between the cities
The total pipe's distance is:
$$Dist=\sqrt{l_1}+\sqrt{l_2}=\sqrt{a^2+x^2}+\sqrt{b^2+x^2}=\sqrt{a^2+\frac{d^2}{4}}+\sqrt{b^2+\frac{d^2}{4}}$$
It doesn't give ##~\sqrt{c^2-4ab}##
 

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  • #2
Orodruin
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You are supposed to minimise the sum of the lengths (i.e., ##L_1 + L_2##) not the sum of the squares of the lengths (##L_1^2 + L_2^2##).
 
  • #3
verty
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Hint: Imagine b is on the other side but d is still the same.
 
  • #4
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You are supposed to minimise the sum of the lengths (i.e., ##L_1 + L_2##) not the sum of the squares of the lengths (##L_1^2 + L_2^2##).
But if the square of the lenghts is minimum then the length is also minimum
I will think about the hint that b is on the other side
 
  • #5
Orodruin
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But if the square of the lenghts is minimum then the length is also minimum
No, this is not true. It is true for a single length, not for the sum of lengths.
 
  • #6
Orodruin
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To expand a bit on that. Note that
$$
(L^2)' = 2LL'
$$
so, for a single length, ##L' = 0## if ##(L^2)' = 0##. However, you have the sum of two lengths and
$$
((L_1+L_2)^2)' = 2(L_1+L_2) (L_1+L_2)' = [L_1^2 + L_2^2 + 2L_1 L_2]' = (L_1^2 + L_2^2)' + 2(L_1 L_2' + L_1' L_2).
$$
Note that ##(L_1^2 + L_2^2)' = 0## no longer guarantees that ##(L_1 + L_2)' = 0## as there are additional terms on the right-hand side.
 
  • #7
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Hint: Imagine b is on the other side but d is still the same.
Capture.JPG

$$\frac{a}{x}=\frac{b}{\sqrt{c^2-(b-a)^2}-x}~\rightarrow~x=\frac{a\sqrt{c^2-(b-a)^2}}{a+b}$$
$$l_1+l_2=...=\sqrt{c^2+4ab}$$
 

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  • #8
verty
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View attachment 229938
$$\frac{a}{x}=\frac{b}{\sqrt{c^2-(b-a)^2}-x}~\rightarrow~x=\frac{a\sqrt{c^2-(b-a)^2}}{a+b}$$
$$l_1+l_2=...=\sqrt{c^2+4ab}$$
You are given a,b,c. Your first task is to find d. Then find L1+L2 when it is the shortest configuration.
 
  • #9
Orodruin
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You are given a,b,c. Your first task is to find d. Then find L1+L2 when it is the shortest configuration.
I think the ##\ldots = \sqrt{c^2 + 4ab}## means he solved it ...
 

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