Find the Ideal Width for a Rectangular Vegetable Garden

[orangeSLICE]

Hi again!

A gardener has a rectangular vegetable garden which is 30m long and 12m wide. He traces to alleys x meters wide, similar to this:

_____________________
|________| x |________|
x___________ _________
|________| x |________|

Determine, with x, the area occupied by the alleys and the area occupied by the vegetables.

Determine the possible values of x for the width of the alley so that the area of the alleys is superior to 41 meters squared and so that the cultivated part is superior to 280 meters squared.

Here's what I've got so far:
30-x * 12 - x = cultivated area
360 - 30x - 12x + x = cultivated area
360 - 42x + x squared = cultivated area
So the gardened area is: x squared - 42x
But how do I prove this? Do I need to prove this?

Next, I need help trying to get the values of x. How do I go about doing this?

Sorry if my "garden" is sloppy. I tried to make it look like the book.

Any help is appreciated Thank you

 

[Fermat]

What's the difference between the cultivated area and the gardened area?

It should be 42x - x² actually, rather than the other way around.

 

[orangeSLICE]

What's the difference between the cultivated area and the gardened area?

There is no difference both names describe the same area. I just don't know how to find the values of the cultivated area and the area of the alleys.

 

[Fermat]

well the area of the cultivated area is 360 - 42x + x².

How did you get 42x - x², or x² - 42x ?

 

[Fermat]

Also the total area of your garden is the sum of the two areas.

You can work out the total area of the garden, yes?

Now subtract the area of the cultivated area to get the area of the alleys.

 

[orangeSLICE]

well the area of the cultivated area is 360 - 42x + x².

How did you get 42x - x², or x² - 42x ?

Um, I'm thinking that 360 is everything cultivated are + alleys. From that I take x² - 42x is the alley area. Am I right?

 

[AKG]

360 is everything, and as you calculated, 360 - 42x + x² is the cultivated area. So the alley area is everything minus the cultivated area. So to find the alley area, it's:

(360) - (360 - 42x + x²)

how are you getting x² - 42x?

 

[orangeSLICE]

360 is everything, and as you calculated, 360 - 42x + x² is the cultivated area. So the alley area is everything minus the cultivated area. So to find the alley area, it's:

(360) - (360 - 42x + x²)

how are you getting x² - 42x?

I have no idea. I just took it from the 360-42x+x².

So I did 360-360+42x-x² and got 42x-x². Now, THAT'S the alley area?

 

[nworm]

360 - 42x + x squared = cultivated area
42x - x squared = area of the alleys
---------------------------------------
360 - 42x + x squared >= 280
42x - x squared >= 41
---------------------------------------
- 42x + x squared >= -80
42x - x squared >= 41
---------------------------------------
x squared - 42x + 80 >= 0
x squared - 42x + 41 <= 0
---------------------------------------
(x - 40)(x - 2) >= 0
(x - 41)(x - 1) <= 0
---------------------------------------
(x <= 2) or (x >= 40)
(1 <= x <= 41)
---------------------------------------
(1 <= x <= 2)

May be the answer is
(1 < x < 2)
I amn't very good in English.
I don't know that means "is sperior to".

 

[orangeSLICE]

i got it

Here's what I found:
-x² + 42x > 41 m
when x = 1
-1 + 42 = 41
so x > 1

360 - 42x + x² > 280 .
x² - 42x > -80
2² - 84 = -80
4 - 84 = -80
-80 = -80
x > 2

Thank you everyone for helping me with this problem! I finally understand it :shy:

 

[Fermat]

Here's what I found:
-x² + 42x > 41 m
when x = 1
-1 + 42 = 41
so x > 1

360 - 42x + x² > 280 .
x² - 42x > -80
2² - 84 = -80
4 - 84 = -80
-80 = -80
x > 2

Thank you everyone for helping me with this problem! I finally understand it :shy:

here's how to do those inequality bits.

You have,

-x² + 42x > 41
or
x² - 42x +41 < 0
which becomes
(x - 41)(x - 1) < 0

Now, if the product of two terms, (x - 41) and (x - 1) is negative, then one of the terms must be -ve and the other term must be +ve.
( -ve means negative and +ve means positive)

let's asume that
(x - 41) is -ve
(x - 1) is +ve
then
(x - 41) < 0 implies x < 41
(x - 1) > 0 implies x > 1
i.e. 1 < x < 41
===========

now assume that
(x - 41) is +ve
(x - 1) is -ve
then
(x - 41) > 0 implies x > 41
(x - 1) < 0 implies x < 1
but x can't be both less than 1 and greater then 41 at the same time, so this assumption is invalid, so only the other assumption is valid.

Hence, 1 < x < 41 for the alley area to be greater than 41 m²
i.e. 1 < x < 41
===========

The other thing you have is, the cultivated area greater than 280 m².

360 - 42x + x² > 280
which becomes
x² - 42x + 80 > 0
(x - 40)(x - 2) > 0

The product of these two terms is +ve, therefore they are both +ve or they are both -ve.

Let's assume both are +ve
(x - 40) > 0 implies x > 40
(x - 2) > 0 implies x > 2
i.e. x must be greater than 40 to make both terms +ve, but maximum size of x can only be 12 m. Therefore this assumption is invalid.

assume that both terms are -ve
(x - 49) < 0 implies x < 40
(x - 2) < 0 implies x < 2
Both terms are -ve for x < 2
Hence, x < 2 for the cultivated area to be greater then 280 m².

Combining these inequalities gives us
1 < x < 2
=======
as the requirement on the size of the alleyways to satisfy the given conditions. viz. that the area of the alleys is greater than 41m² and the area of the cultivated area is greater than 280 m².