[Guest]

A land owner pays a farmer to plough a square field 100m x 100m. The farmer has a special ploughing machine which he can drive round the field in any direction (forwards, backwards, sideways and even diagonally) and can also turn around on the spot. The plough ploughs a square 1m x 1m around its current position.

The land owner says he will pay $0.02 for every square meter he ploughs. Unfortunately, the farmer's plough uses $0.01 of fuel every meter it travels, but it does not use fuel to turn.

What is the most that the farmer can make from this deal?

[Guest]

Hmmmm..... My guess would be

1.11 Because he saves money on the diagonals(one net diagonal per trip around)+ the spot at which he started.

[Guest]

This, for me, is a tough one.

Now, obviously, the minimum he can make is $100. $0.02/m - $0.01/m = 1 cent per meter. There's 10,000 meters, and that would be $100.00. Keep in mind that her earns one cent per meter, not one dollar (just in case someone might make that mistake in the future). Since it takes no gas to turn, He could simply go row by row (or maybe if he wanted to look cool, he could go in a spiral). As well, it's important to note that it can turn around on the spot (I may not know why right now, but someone will). The way to calculate the maximum amount of money would be to incorporate the factor of direction; specifically, a diagonal direction. I don't know how, but there's probably a way to save money on moving diagonally.

Hope this comes in handy for someone else.

[Guest]

I should say that he does not necessarily need to plough the whole field. He will still get payed for what he has done even if he doesn't finish.

Anyway good start so far guys.

[bushindo]

I should say that he does not necessarily need to plough the whole field. He will still get payed for what he has done even if he doesn't finish.

Anyway good start so far guys.

If he moves diagonally, on average, his machine would clear an area of sqrt( 2 )=1.41 m^2 for every meter he travels. That is because if we diagonally move a square that is oriented upright, then diagonal of the square would sweep out a larger area. That would be a net income of 1.41*.02 - .01 = 0.0182 / meter.

So, suppose that he sweeps the field row by row, but orients his machine so that the diagonal line is perpendicular to the direction of travel, he would need to travel roughly 100 * 100/sqrt( 2 ) = 7100 meters roughly, then he needs to pay 71 dollars for fuel. There is little bit of distance travelling from row to row too that I'm not including

He covers the entire field, so he gets paid 200 dollars. That is a net gain of 129 dollars. Refinements of the solution should change the answer a bit, but I expect it to be close to 129.

[plainglazed]

$100.01

If the farmer ever crosses a path that he has already ploughed, it costs him $.01 in fuel. Conversely, regardless of the route he takes, if he never crosses his path, he makes the maximum. The maximum is $.02 x 10,000 sq. whatevers (forgot the measurment) or $200 less the cost of fuel. When he starts out the first square whatever does not cost him $.01 in fuel but the other 9,999 do so his fuel cost is $99.99. Net = $100.01?

[Guest]

lets assume if i can the mower deck is 1 meter square and since it's special and can mow 1 meter square around it, without figuring partial meters, my guess would be...our farmer needs $.31 for fuel, which would net him $1.69. just a little shy of a cold brewsky!

[Guest]

whoops! i forgot i was figuring a 10X10 field with plans of multipling by 100

$169.00...should get a beer or two with that.

lets assume if i can the mower deck is 1 meter square and since it's special and can mow 1 meter square around it, without figuring partial meters, my guess would be...our farmer needs $.31 for fuel, which would net him $1.69. just a little shy of a cold brewsky!

[bonanova]

Since turning costs nothing for fuel, let's assume the farmer constantly spins his machine as it moves along a straight path.

The spinning machine now plows a circular area of diameter 2^.5 = 1.414... meters.

Thus as it moves it plows a path of width 1.414... meters.

To ensure the corners of the square field are plowed,

let the path start with the center of the machine on a boundary [call it side A]

in such a way that the edge of the plowed circle touches a perpendicular boundary [call it side B.]

Plow parallel to B until the center of the machine gets to perpendicular boundary C.

Plow parallel to C a distance of 1.414... meters.

Plow parallel to B again until centered on A.

Plow 1.414... meters parallel to A.

Plow parallel to B again until centered on C.

Continue until the field is covered.

How long is that path?

It's the sum of two lengths:

1. 100 meters x ceiling[100/2^.5] = 100 x 71 = 7100 meters
   That's the length of the paths parallel to side B.
   
   
   
2. 1.414... meters x 71 = 100.41 meters
   That's the length of the paths parallel to sides A and C.
   

The fuel cost is thus $0.01 x 7200.41 = $72.0041.

His profit is $127.99

Owing to the circular shape of the plowed area, he over plows to ensure the corners are covered.

A spiral path might be somewhat shorter, increasing his earnings by a bit.

[Guest]

Since turning costs nothing for fuel, let's assume the farmer constantly spins his machine as it moves along a straight path.

The spinning machine now plows a circular area of diameter 2^.5 = 1.414... meters.

Thus as it moves it plows a path of width 1.414... meters.

To ensure the corners of the square field are plowed,

let the path start with the center of the machine on a boundary [call it side A]

in such a way that the edge of the plowed circle touches a perpendicular boundary [call it side B.]

Plow parallel to B until the center of the machine gets to perpendicular boundary C.

Plow parallel to C a distance of 1.414... meters.

Plow parallel to B again until centered on A.

Plow 1.414... meters parallel to A.

Plow parallel to B again until centered on C.

Continue until the field is covered.

How long is that path?

It's the sum of two lengths:

1. 100 meters x ceiling[100/2^.5] = 100 x 71 = 7100 meters
   That's the length of the paths parallel to side B.
   
   
   
2. 1.414... meters x 71 = 100.41 meters
   That's the length of the paths parallel to sides A and C.
   

The fuel cost is thus $0.01 x 7200.41 = $72.0041.

His profit is $127.99

Owing to the circular shape of the plowed area, he over plows to ensure the corners are covered.

A spiral path might be somewhat shorter, increasing his earnings by a bit.

Looks about right .

I'm not sure about a spiral path, maybe someone could work something out?

[plainglazed]

Of course. Constantly spinning. Well done bonanova. One thought: To make 71 full length say north/south passes (parallel to your side B), doesn't it take just 70 east/west (along either border A or B) jogs?

[plainglazed]

Would not trust my math but I got $128.42 by "spin" ploughing within the 100m x 100m and forgoing the corners and every other whatever the shape is called left by two circles and a line touching. By making 71 vertical paths only 99m long and 70 horizontal paths 1.414m.

Regardless, to me the solution is knowing the plough needs to spin.

[bonanova]

Of course. Constantly spinning. Well done bonanova. One thought: To make 71 full length say north/south passes (parallel to your side B), doesn't it take just 70 east/west (along either border A or B) jogs?

Well, yes, it you don't want to plow the whole field, you'd go 70 times for a total width of 98.99494937 meters.

To maximize the cost effectiveness, that might be just the thing to do, as well as leaving unplowed the corners.

Plowing the 71st row gets you only about 1% additional coverage of the field, nonetheless costs you the full amount in fuel.