Maximum Area

[lindsayvickroy]

Can someone please remind me on how to find the maximum area?

The question was: You have 100 yards of fencing to enlose a rectangular area. Find the dimensions of the rectangle that maximize the enclosed area. What is the maximum area? Width is 50-x; length is x.

Thanks!
Lindsay

 

[arthur ohlsten]

width=50-x x in yds
length=x
area = [50-x]x
area=-x^2+50x by calcxulus this is easy take derivative of area with respect to x, and set it equal to 0, and you get x needed
d[area]/dx= -2x+50 set =0
0=-2x+50
2x=50
x=25

by algebra I would plot -x^2+50x for x from 0 to 50
plot x[50-x] from 0 to 50 the curve will be a maximum at x=25
Arthur

 

[lindsayvickroy]

I really don't understand how that will help me find the maximum area of the rectangle. Can someone else please help me?

 

[tkhunny]

You're almost there.

You have the value of x for which the area is maximum, x = 25 yds. All you have left is calculating the maximum value.

You have Area = x*((50 yds)-x). Maximum area = (25 yds)*((50 yds) - (25 yds)) = 25*25 yds² = 625 yds²