Mellissa has 60ft of fencing for a gardan.She wants a rectangular gardan with the greatest possible area for?

Question

planting.What is the lenth width and area of the gardan she should make?(use only whole units for the lenth and width.PLease explain how to get answer

Answer 1

If she did a square she would have the biggest area. That would make all the sides 60/4 = 15 feet long. If she really wants a rectangle not a square and only whole numbers she would need to make the short sides 14 feet long and the long sides 16 feet long. That gives her the same total of 60 feet of fencing with an area of 224 sq feet. The square would have been 225 sq feet.

Answer 2

I'm assuming you are in Calculus, if not ignore this comment.

P=2L + 2W
A=L*W

First set up an equation for the perimeter of the garden.
60=2L+2W

Solve for W or L
W=(60-2L)/2

Plug in to the Area equation
A=L*((60-2L)/2)

Differentiate this equation then find x=0. This will give you the value of the length. Then plug the length back into the perimeter equation to find the width value. Finally, multiply the width and length together to get the maximized area.

Answer 3

optimal area exist whilst area length cases end length is optimal and the area around the rectangle is 28 yards. initiate with a element length of 13 and a end length of a million and calculate the area. In increments of one improve the tip length by technique of a million jointly as reducing the realm length by technique of a million till the realm will become a million and the tip length becomes13. Calculate the area after each and every incremental replace. the optimal area will ensue whilst the realm length is 7 and the tip length is 7. the optimal area would be 40 9 sq. yards.

Answer 4

15' x 15'
Area = 225'
A square is always the rectangle with the greatest area for a given perimeter.
Try 16' x 14'
Same perimeter (60')
Area = 224'

Answer 5

It's a trick question. Squares ARE rectangles, they're just rectangles with 4 equal sides. I believe the problem is simply specifying that you are not allowed to build a circle with a 60ft circumference, or a triangle with 3 20ft sides, etc.
The answer is 15 x 15.

Answer 6

2x+2y=60, with x being the width and y being the length
2y=60-2x
y=30-x

a minimum or maximum occurs when the first derivative of the wanted variable (area in this example) is equal to zero
Area=xy
A=x(30-x)
A=30x-x^2
A'=30-2x=0, where A' is the first derivative of Area
30=2x
x=15ft

Back to the first equation
y=30-x
y=30-15=15ft

ANSWER:
length: 15ft
width: 15 ft
If you can't have a square, make one side (width) slightly larger and the other (length) slightly larger--perhaps 14ft and 16ft

Answer 7

16ft. by 14ft. = 224sq. ft

As area is calculated by multiplying length x width, the greatest possible area is a square (15ft by 15ft = 225sq ft), but as it has to be a rectangle, increase one side by a foot and decrease the other by a foot to reach 60 total feet with the greatest possible area (i.e. 16ft +16ft+14ft+14ft =60ft).

Answer 8

I believe its 14 x 16 which would have an area on 224, but I didnt work to hard on it either...

Answer 9

the answer is 14 ft by 16 ft.