The numbers are so easy you dont need algebra as a first guess works:
The garden is 5x9 so area is 45
The border are is 24
Therefore the centre area is 45-24=21
If the problem maker is a nice guy he will make the solution in whole numbers so lets try a central area of 3x7
That means border is 1 foot.
Check it out
Originally posted by wolfgang59obviously right, so here is the algebra
The numbers are so easy you dont need algebra as a first guess works:
The garden is 5x9 so area is 45
The border are is 24
Therefore the centre area is 45-24=21
If the problem maker is a nice guy he will make the solution in whole numbers so lets try a central area of 3x7
That means border is 1 foot.
Check it out
45 -(4x^2 - 28x + 45) = 24 The trinomial represents the area of
the inside rectangle and x equals
the width of the border
yada yada
two solutions but 1,6 6 is to large so x = 1
so there is the algebra that wasnt needed.....hahah
Originally posted by tournymangrWith just the info you have supplied there is no unique answer.
ok new one!
....6.....|.....8.....|...11....|...15
....4.....|.....7.....|...5......|...?
8......5..|.17...5.|..6....11.|.8.....6
find the ? by using the same relationship as the other triangles
If we assume something like
F(x, y, z) = w and is a linear function of x,y,z
Where x is top number and y & z are bottom numbers. w is middle number
I think you get your unknown as 169/21 .. but thats on a small scrap of paper I will check but doubt that is answer you want!!!
Originally posted by wolfgang59uhhhh.....no. if you multiply the bottom numbers then subtract the top, you get the centre number, twice. i.e 17*5=85, 85-8=77, the centre number would be 7.
With just the info you have supplied there is no unique answer.
If we assume something like
F(x, y, z) = w and is a linear function of x,y,z
Where x is top number and y & z are bottom numbers. w is middle number
I think you get your unknown as 169/21 .. but thats on a small scrap of paper I will check but doubt that is answer you want!!!
Originally posted by tournymangrOK
uhhhh.....no. if you multiply the bottom numbers then subtract the top, you get the centre number, twice. i.e 17*5=85, 85-8=77, the centre number would be 7.
so your solution is
F(x,y,z) = 11(yz-x)
tidier than mine but no more valid. As I said there are an infinite number of solutions depending on how devious you want to be.
My linear solution is unique. Your quadratic solution isnt.
Originally posted by wolfgang59yes your solution does give mine a kick in the butt but it wasnt meant to be so hard. the formula you came up with is genius. i hadnt thought it that way before. well done!
OK
so your solution is
F(x,y,z) = 11(yz-x)
tidier than mine but no more valid. As I said there are an infinite number of solutions depending on how devious you want to be.
My linear solution is unique. Your quadratic solution isnt.