Thread: help with the most economical pipe.....

1. help with the most economical pipe.....

The cost, £P million, of laying one kilometre of pipe for a water main(=supply) is calculated by means of the formula

P=4000/9a+4a

where a is the cross-sectional area of the pipe in square metres.

a: What is the cross-sectional area of the most economical pipe to use? (Answer to the nearest tenth of a square metre)

b:Calculate the minimum cost of laying one kilometre of pipe (Answer to the nearest £ million)

I do not want to discuss million or billion.....
English homework is somewhere else!

to a: I know I have to work with the derivative of the function
P(a)=4000/9a+4a

P^1(a)=4000/9a^-1+4a^-1 next P^1(a)=0

4000/9a^-1+4a^-1=0 then I'm lost

2. Originally Posted by Eenhoorn
The cost, £P million, of laying one kilometre of pipe for a water main(=supply) is calculated by means of the formula

P=4000/9a+4a

where a is the cross-sectional area of the pipe in square metres.

a: What is the cross-sectional area of the most economical pipe to use? (Answer to the nearest tenth of a square metre)

b:Calculate the minimum cost of laying one kilometre of pipe (Answer to the nearest £ million)

I do not want to discuss million or billion.....
English homework is somewhere else!

to a: I know I have to work with the derivative of the function
P(a)=4000/9a+4a

P^1(a)=4000/9a^-1+4a^-1 next P^1(a)=0

4000/9a^-1+4a^-1=0 then I'm lost
$P = \frac{4000}{9 a} + 4a = \frac{4000}{9} a^{-1} + 4a$

$\frac{dP}{da} = (-1) \frac{4000}{9} a^{-2} + 4 = - \frac{4000}{9 a^2} + 4$ using the standard rule.

Your job is to solve $0 = - \frac{4000}{9 a^2} + 4$ and substitute the answer into the formula for P.

3. Originally Posted by mr fantastic
$P = \frac{4000}{9 a} + 4a = \frac{4000}{9} a^{-1} + 4a$

$\frac{dP}{da} = (-1) \frac{4000}{9} a^{-2} + 4 = - \frac{4000}{9 a^2} + 4$ using the standard rule.

Your job is to solve $0 = - \frac{4000}{9 a^2} + 4$ and substitute the answer into the formula for P.
I should add that there is more than one solution. A job to add to your list is to determine the nature of each solution. Obviously you want the solution that corresponds to the minimum turning point (you will no doubt find that this is the positive solution for $a$. The negative solution will correspond to a maximum turning point).