Question

If you have 120 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

\(2x+2y=120\)
and
\(f(x) = xy\implies f(x)=x(60-x)\implies f '(x)=60-2x\)

Answer 2

with me?

Answer 3

yea I think so

Answer 4

perimeter is given by p=2x+2y and they told us we have 120 to use
so 120=2x+2y this is the same as 60=x+y which implies y = 60-x

now area is given by A = xy but y = 60-x so we sub n for y

A=x(60-x)

this is what we want to maximize, so we take the derivative and get

f'(x) = 60-2x

right?

Answer 5

that should say sun in not "sub n"

Answer 6

ok now we set the derivative equal to 0

60-2x=0 which gives x=30

so we need to find y, so we use 2x+2y=120 which is again the same as x+y = 60

if x is 30 we know y is also then 30 , so our max area is 30*30=900

Answer 7

Thank you so much, I just could not figure out how to set it up. that was very helpful!

Answer 8

Oh my gosh, so if you are using 120 meters and we found the largest area to enclose what are the units for that?

Answer 9

m^2?

Answer 10

Yes, the units of measurement of area in this case would be m^2 (meters squared).