Question

The perimeter of a paved area being built at a playground is going to be 46 meters. If the length and width are each a whole number of meters, what should the length and width of the area be to cover the greatest possible area? Draw a picture and make an organized list to solve.

Answer 1

Drat...stand by, I screwed this one up in my head already.

Have to think a bit...

Answer 2

2w + 2L = 46 div everything by 2

w + L = 23 now make a list of the possibilities
1 22 area = 22
2 21 area = 42 keep going until you find largest area

Answer 3

does that make sense?

Answer 4

the perimeter can only be 46 though

Answer 5

This is true.

Which of my examples didn't have a perimeter of 46 units?

Answer 6

DAMMIT!!!! You're correct!!!

I based my crap on 64 units, not 46!!!!!

Answer 7

P = 2W + 2L
2(1) + 2(31) = 64

Answer 8

My apologies, and thank you for pointing it out!!

Answer 9

I'm going to delete my 2 offending posts and start over...

Answer 10

10* 13 = 130
11*12 = 132, I think this would be the biggest area

Answer 11

Yes, agreed.

11*12 (or 12*11) is about as close to a square as you're going to get using integer-length sides. :-)

I just hope I didn't cause jacke any problems with my previous ramblings... :(

Good show, Jim!! :-)

Answer 12

its all good...

Answer 13

Taking Jim766 equation w + l= 23
Then l=23-w
Area = lw or w(23-w) or -w^2 + 23w
Now differentiate and set to 0 -2w + 23 =0 multiply thru by -1/2
w=11.5 then l=23-11.5=11.5 resulting in a square 11.5 by 11.5

Answer 14

let x be the length & y be the width of the park then
P=2(x+y)=46
x+y=23
y=23-x
Area A =x*y=x(23-x)=23x-x^2
dA/dx =23-2x
dA/dx=0, gives 23-2x=0,2x=23
x=23/2
d^2A/dx^2=-23
at x=23/2, d^2/dx^2 is negative
Hence x=23/2,y=23-23/2=23/2
But x &y are whole numbers & perimeter is 46
Hence length =12 & width =11