
Nestor R. answered 07/12/19
Statistician with a very good grounding in Algebra
Let L1, L2 and L3 be the 3 sides requiring fencing. L1+L2+L3 = 40.
Let L2 = L3, so the above equation can be rewritten as L1 + 2L2 = 40 (1)
L1*L2 = 192 = area of the enclosed plot. (2)
Solve (1) for L1:
L1 = 40 - 2L2.
Plug this into (2) to get (40 - 2L2) * L2 = 192 ==> 40L2 - 2L22 = 192. (3)
Rearrange terms of (3) to get 2L22 - 40L2 + 192 = 0
Divide both sides by 2 to get L22 - 20L2 + 96 = 0
Use the quardratic equation to find 2 solutions for L2 = 12 and 8
When L2 = 12, L1 = 16. Since L2 = L3, L1+L2+L3 = 16+12+12 = 40.
12 x 16 = 192.
12 and 16 satisfy conditions for length of fencing and area.
Now try L2 = 8. L2 = L3, so L1 = 40 - 2*8 = 24
24 x 8 = 192.
24 and 8 satisfy conditions for length of fencing and area.
Thus there are 2 possible solutions to the problem.