let the width be x m
let the length be y m
So we need 2 widths and 1 length
given: xy= 448
y = 448/x
2x + y = 60
2x + 448/x = 60
times x
2x^2 + 448 = 60x
x^2 - 30x + 224 = 0
(x-16)(x-14) = 0
x = 16 or x = 14
then y = 448/16 or y = 448/14
y = 28 or y = 32
The field could be 16 m wide and 28 m long, (the 28 m side parallel to the barn)
or
The fields could be 14 m wide and 32 m long, ( the 32 m side parallel to the barn)
A man uses 60m of fencing to make 3sides of a rectangular fence,the fourth being the wall,if the area enclosed is 448m^2 .Find the possible length of the sides of the fence.
1 answer
1 answer