err! ok math problem. "a farmer wants to enclose three sides of a rectangular area that borders a creek. He has 2400 meters of fencing material. What is the max area that can be enclosed by the fence?" now im supposed to use quad equations right now. and the line of symmetry and vertex have something to do with it. how the heck do i answer this!?
2006-12-02 23:08:37 · 8 answers · asked by tejadasammy 2 in Science & Mathematics ➔ Mathematics
ok.... it has to do with a QUADRATIC EQUATION! its not basic algebra thats why im stuck!! please help!!
2006-12-02 23:14:53 · update #1
quadratic... ya know... like ax^2 + bx + c
the formula for finding line of symmetry is
-b/2a
2006-12-02 23:30:08 · update #2
Sum of the length of three sides=2400
That is 2l + b =2400
b = 2400- 2l
Area A = l * b =l *(2400 - 2 l)
= 2400 l -2 l^2
When the area is maximum dA/dl =0
ie. 2400 - 4 l = 0
4 l = 2400
l = 600
b = 2400 - 2*600
=1200
A = 1200* 600 =720000 sq. m
2006-12-03 00:16:14 · answer #1 · answered by ATS 2 · 0⤊ 0⤋
Let x be the variable of the lengths of the rectangle and y be the side of the rectangle.
Then you have if you only enclose three sides of the rectangles
2*x+*y= 2400 <=>
y=2400-2*x <=>
The area of the rectangle, will then be
A=x*y= x*(2400-2*x) =2400x-2x^2
In order to find the maximum area then we have to differentiate with respect to x, and we get:
dA/dx=2400-4x =>
dA/dx= 0 => x=600 [m]
And y will bee y= 2400-2*600 = 1200
So the maximum area will be
Amax= x*y =1200* 600 = 72000 [m^2]
===========================
2006-12-03 07:52:58 · answer #2 · answered by Broden 4 · 0⤊ 0⤋
Use optimization
Area = l*w
Perimetre = 2l + w ( since one of the sides is the creek u can take off one w or l)
Substituting and rearranging
1200 = 2l+w
(1200 - w) /2= l
Area = (1200 - w)w /2
= 600w-w^2/2
Now we have a formula we have to derive it find the second derivitive to dertermin whether its a max or min f''(x) > 0 = min, f''(x) < 0 = max
f(x) =600w - w^2/2
f'(x) = 600-w
f''(x) = -2 .'. max
Now put the first derivitive = to zero
600-w = 0
w = 600
2400 = 2l+w
1800 = 2l
l = 900
.'. The max area = 900*600= 54000m^2
2006-12-03 07:47:30 · answer #3 · answered by Anonymous · 0⤊ 0⤋
I didn't use algebra but i used calculus to figure out the maximum area. if the farmer were to make the to sides of the fence 600 meters long and then the other longer side 1200 meters long this would give him the maximum amount of area.
A= l * w
600 * 1200= 720,000 m2
2006-12-03 07:25:52 · answer #4 · answered by ? 2 · 0⤊ 1⤋
the area of a square is the largest area of all kinds of rectangles with a certain perimeter
the perimeter is 2400 meters so each side is 600 meters
the area is : 600^2 = 360,000 m^2
2006-12-03 07:26:18 · answer #5 · answered by James Chan 4 · 0⤊ 1⤋
max area is when the lenghts of the sides are equal, if he only needs to fence 3 sides then each length will be 2400/3 = 800m
therefore max area is 800 x 800 = 640000 metres squared
easy!
2006-12-03 07:13:14 · answer #6 · answered by Anonymous · 0⤊ 1⤋
use calculus mate
let x = be one side of the rectangle
2400-x= the other side
A = x(2400-2x)
dA/dx = 2400 - 4x
at maximum dA/dx = 0
0 = 2400 - 4x
x = 600 meters
A(max) = 600 (2400 -1200)
= 720,000 sq. meters
2006-12-03 09:08:55 · answer #7 · answered by para_tubag 1 · 0⤊ 0⤋
a +2b = 2400, ab ==> max a=2400-2b so (2400b-2b^2) max b 0,2400 f=2400b-2b^2 f'=0=2400-4b so b=600 note as area is + 0==>0 a2400==> - thus lenths = 600, 1200, 600
2006-12-03 07:39:04 · answer #8 · answered by mathman241 6 · 0⤊ 1⤋