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

## Homework Equations

Minimum/Maximum occurs when the first derivative=0

## The Attempt at a Solution

$$l_1^2+l_2^2=(a^2+x^2)+[b^2+(d-x)^2]=a^2+b^2+x^2+\left[ \sqrt{c^2-(b-a)^2}-x \right]^2$$
$$(l_1^2+l_2^2)'=2x+2\left[ \sqrt{c^2-(b-a)^2}-x \right](-1)$$
$$(l_1^2+l_2^2)'=0~~\rightarrow~~x=\frac{1}{2}\sqrt{c^2-(b-a)^2}=\frac{d}{2}$$
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 $~\sqrt{c^2-4ab}$

#### Attachments

• 24.7 KB Views: 364
• 5.2 KB Views: 378

Related Calculus and Beyond Homework Help News on Phys.org
Orodruin
Staff Emeritus
Homework Helper
Gold Member
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 ($L_1^2 + L_2^2$).

verty
Homework Helper
Hint: Imagine b is on the other side but d is still the same.

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

Orodruin
Staff Emeritus
Homework Helper
Gold Member
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.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
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.

Hint: Imagine b is on the other side but d is still the same.

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

#### Attachments

• 6.6 KB Views: 282
verty
Homework Helper
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.

Orodruin
Staff Emeritus
I think the $\ldots = \sqrt{c^2 + 4ab}$ means he solved it ...