# Mr. Young has two rectangular trays with side lengths that are whole numbers. The first tray has an area of 100 square inches and a perimeter of 50 inches. The second tray also has an area of 100 square inches, but with a smaller perimeter. Find the perimeter of the second tray in inches. Show your work using words or numbers.

1
by soldierxtra256

2014-10-31T21:02:59+08:00
For the first tray:
let 'x' be the length of one side
'y' be the length of the other side
for area:
xy = 100        ----equation 1
for perimeter:
2x + 2y = 50 ----equation 2
dividing the equation 2 with 2 you'll have:
(2x + 2y = 50) /2
x + y = 25
transpose y to the right side of the equation
x = 25 - y  -----equation 2'
substitute equation 2' to equation 1
xy = 100
(25-y)(y) = 100
distribute y  to the quantity (25-y)
25y - y² = 100
transposing 25y - y² to the right side, equating it to zero you'll have:
y² - 25y + 100 = 0
factoring you'll have:
y² - 25y + 100 = 0
(y - 20)(y - 5) = 0
y - 20 = 0       y - 5 = 0
y = 20             y = 5
substitute the values to equation 2'
x = 25 - y
when y=20
x = 25 - 20
x = 5
when y=5
x = 25 - 5
x = 20

for the second tray:
let 'x' be the length of one side
'y' be the length of the other side
for area:
xy = 100   ----equation 1
for perimeter:
2x + 2y < 50    -----inequality 1
since we are required to have the dimensions in whole number and that the perimeter should be less than 50 then we can have the dimensions as the factors of 100:
FACTORS of 100
100: -1 and 100
-2 and 50
-4 and 25
-5 and 20
-10 and 10
since in this case, the perimeter is to consider such that its perimeter would be less than 50 then we'll check for the factors of 100 that will give as the required perimeter.
Perimeter:
for factors 1  and 100
P = 2x + 2y
P = 2(1) + 2(100)
P = 2 + 200
P = 202 > 100 not ok!
for factors 2 and 50
P = 2(2) + 2(50)
P = 4 + 100
P = 104 > 50 not ok!
for factors 4 and 25
P = 2(4) + 2(25)
P = 8 + 50
P = 58 > 50 not ok!
for factors 5 and 20
P = 2(5) + 2(20)
P = 10 + 40
P = 50 ; since this isn't less than 50 then this is not ok!
for factors 10 and 10
P = 2(10) + 2(10)
P = 20 + 20
P = 40 < 50 ok!
therefore you'll have the dimensions of the second tray as
10inches by 10inches