Question

Alex has 520 yards of fencing to encolose a returangular area.

A retaungle that maxmises the enclosed area has a length of _ yards and a width of _ yards.

The maxium area is _____ yards

?

Answer 1

perimeter:
p=2(w+L)=520
w=260-L
A=wL=260L-L^2
at max
dA/dL=260-2L=0
L=130
w=260-L=130
A(130)=130^2=16900m^2

Answer 2

huh

Answer 3

note that the maximum area is found when w=L i.e. a square maximizes the area of a rectangle.

Answer 4

what math are you in kandice?

Answer 5

college algebra

Answer 6

I don't get how to get the maxium... thats where im stuck

Answer 7

well in that case ignore the dA/dL part because that's calculus
so look at it this way:
We are told the perimeter is 520. The formula for perimeter is
p=2w+2L=520.
The only way for you to solve this is by recognizing that the area is maximized by a square, so L=w. Now the above becomes
p=2w+2w=520=4w
w=130
now just find the area
A=wL=130(130)=16900

Answer 8

so the max would be 16900

Answer 9

There is no way I know of to find the maximum of a given function without calculus, except by recognizing a special case like this from prior knowledge.

Answer 10

yes that is the maximum area possible

Answer 11

when w=L
i.e. a square

Answer 12

by the way the maximum area is given in yards-squared, not yards as it says above.