HELP (pre calc problem)?

Question

I know the solution is going to be really obvious, but I need help!

Okay, so there is a field with an area of 2000. Using the derivitive to help solve it, how can I find the max amount of fencing to put around the field?

so far, I have
2000 = lw
w = 2000/L
2L + 2000/L
2L + 2000L^-1
' = 2 - 2000L^-2
4 - 2000L = 0
L = 500.... but this is wrong... help?!

maximum.....

shoot, i'm stupid... mimimum abount of fencing :)

Answer 1

Max or min?

The "max" amount of fencing for a "field of area 2000" is infinite.

Imagine the field is a rectangle whose width is 0.001. Then the length would be A/W (2000/0.001) = 2,000,000, to have an area of 2,000. The amount of fencing needed would be 4,000,000.002.

And if the width were 10 times smaller yet (0.0001) then the length would have to be ten times larger yet (20,000,000) to still have 2,000 area... and the amount of fencing would be 40,000,000.0002.

Now, assuming that you meant the minimum amount of fencing, and that the field is a rectangle, let's solve that problem:

A = LW
2000 = LW
W = 2000/L

The perimeter of a rectangle is:

P = 2L + 2W
P = 2L + 2(2000/L)
P = 2L + 4000/L

To find the minimum perimeter, we take the derivative with respect to L, and set that equal to zero:

dP/dL = 0
d(2L + 4000/L)/dL = 0
2 - 4000/(L^2) = 0
4000/(L^2) = 2
4000 = 2L^2
L = sqrt(2000)

A = LW
2000 = sqrt(2000)W
W = 2000/sqrt(2000)
W = sqrt(2000)

So, the minimum fencing (perimeter) for a rectangular field of area 2,000 is used when the width and length are each sqrt(2000) -- it's a square, with side sqrt(2000) = 20 * sqrt(5).

The fencing (perimeter) of a square is four times the length of its side, 80 sqrt(5).

Answer 2

a million] you should eliminate between the unknowns with the help of substitution using 2d equation y = x^2 -a million placed this fee for y contained in the first equation 11x - 2[x^2 -a million] = -4 11x - 2x^2 + 2 = -4 2x^2 - 11x - 6 = 0 [2x + a million][x -6] = 0 x = + 6 or - a million/2 if x = -a million/2 y < 0 So x should be = +6 This makes y = 35 2] using lengthy branch [x + a million][4x + 2] = 4x^2 + 6x + 2 yet we are instructed there's a the relax 2 this provides 4x^2 + 6x + 2 + 2 = 4x^2 + 6x + 4 So L = 4

Answer 3

I think a little more information is required. Do you really want the maximum amount of fencing or the minimum? Minimum would make more sense (who wants to spend the most possible). Usually, these problems have another constraint. Have you told us everything?

Answer 4

area: 2000 = xy
perimeter=2x+2y

2000/x=y
substitute in perimeter

2x + 2(2000/x) = p

differentiate that and hopefully it should be the right solution :)