Question

If you have 240 meters of fencing and want to enclose a rectangular area up against a long, straight wall, what is the largest area you can enclose?

Answer 1

3x + y = 240
xy is a max

use lagrange multipliers

Answer 2

well, 2x + y = 240, thought it was a different problem

Answer 3

It doesn't have to be that complicated I think... Think what kind of rectangle will give you the biggest area.

Answer 4

granted there are other ways to approach it :)

Answer 5

and by lagrange multipliers are you talking about the thing that you learn in calculus? O.o

Answer 6

yes

Answer 7

What are those?

Answer 8

Eh this is a kind of a question you would get in a grade 9 math xD

Answer 9

Thats funny considering my calc 1 class is asking it

Answer 10

if memory serves :)

f(x,y) = xy
g(x,y) = 2x + y - 240


fx = y, Lgx = 2L
fy = x, Lgy = L

x = L, y = 2L

2(L) + 3(2L) = 240

8L = 240, L=30

x = 30, y = 60

Answer 11

typoed in the middle ...

Answer 12

Thank you got it :)!

Answer 13

2x + y = 240

2(L) + (2L) = 240

L = 60

Answer 14

I dont know lol I got that question in my grade 9 math test lol I remember

Answer 15

letting the area be defined as: xy

and the constraint as: 2x+y = 240

we see from the constraint that y = 240-2x

x(240-2x) is an upside down parabola with the highest point at x=60

Answer 16

which would be the algebra way .... :)

Answer 17

Let x and y be the sides of rectangle, then
2x+y=240
y=240-2x
\[Area A=xy=x(240-2x)=240x-2x ^{2}\]
\[\frac{ dA }{ dx }=240-4x\]
\[\frac{ dA }{ dx }=0,240-4x=0,4x=240,x=\frac{ 240 }{ 4}=60\]
\[\frac{ d ^{2}A }{ dx ^{2} }=-4\]
\[at x=60, \frac{ d ^{2}A }{dx ^{2} }=-4<0\]
Hence A is maximum at x=60
and y=240-2*60=240-120=120
Hence it is asquare with each side=120 m

Answer 18

not a square