A farmer wants to make a rectangular field with a total area of 3200. It is surrounded by a fence. It is divided into 3 equal areas by fences. What is the shortest total length of fence this can be done with.
2007-01-23 19:10:53 · 7 answers · asked by Mike K 1 in Science & Mathematics ➔ Mathematics
<>A square will give you the shortest perimeter, so the sq root of 3200 is 56.569. With 4 sides, and 2 additional dividers your final answer is 6 * 56.569 or 339.411 of whatever you are measuring in.
2007-01-23 19:28:59 · answer #1 · answered by druid 7 · 0⤊ 2⤋
Draw a picture of three equal size rectanges side by side.
If we assume that the fences are arranged as above, then the problem is not too hard. Let the height be h and the width be w.
The total fencing needed is then T = 4*h + 2*w. Solving for w we get
w = (T - 4 * h) / 2.
The area is
A = h * w
= h * (T - 4 * h) / 2
= - 2 h^2 + (T / 2) * h.
So A is a nice parabola. For a parabola with formla
a h^2 + b h + c
the peak occurs at h = - b / (2 a). So the peak of the A parabola is at
h = - (T/2) / (2 * (-2)) = T/8.
w = (T - 4 * h) / 2 = (T - 4 * T/8) /2 = T/4.
A = h * w = T/8* T/4 = T^2 /32.
A = 3200, so
3200 = T^2 /32
32*32*100 = T^2
T = 320, h = 40 , w = 80.
The total length of fencing is 320.
2007-01-24 03:43:01 · answer #2 · answered by Hans S 1 · 1⤊ 1⤋
the Area is :
xy = 3200
the Perimeter is:
P = 4x + 2y
use the area formula to remove y from the perimeter formula:
y = 3200/x
P = 4x + 6400/x
Since Hans already posted the Pre-Calc approach i'll post the calc approach.
take the derivative of P and set equal to zero:
P' = 4 - 6400/x^2
x = 40 (this is the onre we want on a physical basis)
or
x=-40
so plug x = 40 into the perimeter and you're done:
P = 4*40 + 6400/40
P = 160 + 160
P = 320
2007-01-24 04:19:10 · answer #3 · answered by cp_exit_105 4 · 1⤊ 0⤋
a = s1*s2
per = s1+s2+s3+s4 -- since s3=s1 and s4 = s2 we make it
per - 2*(s1+s2)
50*64 would be an area of 3200 and a perimieter of 2*(50+64)=2*(114) = 228 ..... and then we would need to run two 50 ft fences at a spot 21 1/3 ft on the 64 foot wall and one at 42 2/3 ft as well. this divides the rectangle in 3
So this uses 228 feet for the perimiet plus 100 more for the dividers
As rectangles get close to squaresm less is needed for the perimieter and but more is needed for the dividers
2007-01-24 03:29:12 · answer #4 · answered by Bill F 6 · 0⤊ 2⤋
OK...I get it now.
A*B = 3200
B = 3200/A
minimize 4A + 2B
Since B = 3200/A
minimize 4A + 6400A^-1
Find the derivative:
4A^2 - 6400 = 0
A^2 - 1600 = 0
A=40
B=90
Total fence = 4*40 + 2*90 = 320
2007-01-24 03:39:44 · answer #5 · answered by gebobs 6 · 0⤊ 1⤋
Can you tell me if it's 3200 metres, kilometres, inches, or centimeteres?
Your question makes no sense.
2007-01-24 03:19:57 · answer #6 · answered by Anonymous · 0⤊ 2⤋
3200 km,m, cm,what???
2007-01-24 03:28:28 · answer #7 · answered by Az 3 · 0⤊ 2⤋