Question

A rancher wants to fence in an area of 2693400 square feet in a rectangular field and then divide it in half with a fence down the middle parallel to one side. What is the shortest length of fence that the rancher can use?

Answer 1

hmm.. the "fence down the middle" confuses me.... but here's what I think..
The minimum perimeter is a square...
so the length = sqrt (2693400) = 1641.158128
There are a total of 5 equal lengths of the fence due to the middle fence... so the total fence = 1641.158128 x 5
= 8205.790638
Just round it to whatever your teacher needs...

Answer 2

You have to take into account the fence down the middle.

Answer 3

yeah, that's what confuses me...

Answer 4

is the fence down the middle just the same as one of the width of the rectangular fence

Answer 5

Yes.

Answer 6

Set up an equation for the area of the field. Set that equal to 2693400. Set up the equation for the perimeter of the fence. Use the equation of the area to make the perimeter equation have only one variable instead of 2. Use the derivative to find the local mins of the perimeter function.

Answer 7

ok thanks i will do that if i get stuck i will ask for more help :) but that seems to clarify things

Answer 8

If you get stuck, post here what equations you're using for area and perimeter.

Answer 9

A=LW; P=2L+2W I get P=2L+2(2693400/L)

Answer 10

That perimeter is not quite right because you need to take into account the fence down the middle. So P = 2L + 3W

Answer 11

ok yea i don't know why i am leaving it out lol

Answer 12

so my P'=2+8050200/L^2

Answer 13

Close, but
\[ \frac{d}{dx} (\frac{1}{x}) = \frac{-1}{x^2}\]

So you've got a sign problem in P'

Answer 14

yea i saw it so i get 2/8050200=L^2 after i fix sign error then i get sqrt(2/8050200)=L

Answer 15

uh.. I think you're playing a bit fast/loose here..

\[P' = 2-\frac{8050200}{L^2}\]
\[P'=0 \rightarrow \frac{8050200}{L^2} = 2\]
\[\rightarrow \frac{8050200}{2} = L^2 \]

Answer 16

You have it the other way around.

Answer 17

yea cause its like saying 8050200(1/L^2)

Answer 18

ok so will L be my answer..

Answer 19

I am getting a wrong answer...

Answer 20

The L you find by setting P' to 0 will be the length at which your perimeter function will be at its minimum value. You then need to plug in this L to your normal perimeter function to find how much fence the farmer needs.

Answer 21

my math is wrong it was 8080200 not 8050200

Answer 22

I got the right answer...thanks for your help sorry i was just making minor errors but you helped me see them

Answer 23

could you help me with another
If 3072 square centimeters of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

Answer 24

Same type deal. Set up equation for surface area, and volume. Eliminate a variable, take the derivative of the volume function and find a local max (where V' = 0). Plug in that dimension back into V to find the max Volume

Answer 25

err 2 equations, one for surface area, one for volume.

Answer 26

is it surface area because we know the amount of material that we use

Answer 27

it's surface area because they give us the material yes (in square centimeters)

Answer 28

V=w^2*H and my surface area would be of a cube?

Answer 29

Yes with the top missing.

Answer 30

so its just five sides SA=5x^2 or is it SA=2w^2+4wh

Answer 31

That last one. The length and width is the same because the bottom is square, but it doesn't stipulate that the sides are square. Although it should only be w^2 not 2w^2 since the top is open.

Answer 32

ok

Answer 33

i am kinda lost with the derivative...

Answer 34

What's the equation you have for Volume?

Answer 35

V=w^2(3072-w^2/4w)

Answer 36

Shouldn't that be
\[ V = \frac{w^2(3072-w^2)}{4w}\]

Answer 37

could you eliminate the w from the bottom and one w from w^2

Answer 38

yes.

Answer 39

\[ V=\frac{1}{4} * [3072w - w^3] \]

Answer 40

V'=1/4[3072-3w^2]

Answer 41

yep.

Answer 42

Thanks again i got it again... i am starting to understand optimization

Answer 43

A rectangle is inscribed with its base on the x -axis and its upper corners on the parabola
y=2-x^2

What are the dimensions of such a rectangle with the greatest possible area?
width=
height=
i figured out height was 4/3 cause i thought width was \[\sqrt{2/3}\] and plugged it in y=2-x^2