Mark’s garden is a rectangle that is (4a+96)/(a+15) meters long and (4a+24)/(a+15) meters wide. If Mark decides to fence his garden, what would the total length of fencing material required be?
2 ((4 a+96)/(a+15)+(4 a+24)/(a+15))
(4 a+96)/(a+15) = 96/(a+15)+(4 a)/(a+15):
2 (96/(a+15)+(4 a)/(a+15)+(4 a+24)/(a+15))
(4 a+24)/(a+15) = 24/(a+15)+(4 a)/(a+15):
2 (96/(a+15)+(4 a)/(a+15)+24/(a+15)+(4 a)/(a+15))
Grouping like terms, 96/(a+15)+(4 a)/(a+15)+24/(a+15)+(4 a)/(a+15) = ((4 a)/(a+15)+(4 a)/(a+15))+(96/(a+15)+24/(a+15)):
2 (((4 a)/(a+15)+(4 a)/(a+15))+(96/(a+15)+24/(a+15)))
4 a/(a+15)+4 a/(a+15) = 8 a/(a+15):
2 ((8 a)/(a+15)+(96/(a+15)+24/(a+15)))
96 1/(a+15)+24 1/(a+15) = 120 1/(a+15):
2 ((8 a)/(a+15)+120/(a+15))
2 (120/(a+15)+(8 a)/(a+15)) = (2×120)/(a+15)+2×(8 a)/(a+15):
(2×120)/(a+15)+(2×8 a)/(a+15)
2×120 = 240:
240/(a+15)+(2×8 a)/(a+15)
2×8 = 16:
Answer: | 240/(a+15)+(16 a)/(a+15)
NO MATTER WHAT THE VALUE OF a, THE MAXIMUM PERIMETER WILL ALWAYS BE 16.
