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Get access to the detailed solutions to the previous years questions asked in IIM IPMAT exam
Let BD = x, BF = y
We have to find maximum value of area of rectangle BDEF i.e. maximum value of x.y.
Let the area of rectangle BDEF, A = x ⋅ y …... (i)
?AFE ~?ABC (AA criteria)
∴ AF/AB = FE/BC
(18 − y)/18 = x/18
y = x − 18 .….. (ii)
Putting y = x − 18 in eqn. (i), we per–
A = x(x − 18)
A = x2 − 18x
Differentiating A w.r.t x, we get dA/dx = 2x − 18
Equating dA/dx to 0, in order to get the value x for which area,
A will be maximum, we get
dA/dx = 2x − 18 = 0
x = 9
∴ x = y = 9
∴ Maximum area of rectangle BDEF = x.y
= 9 × 9 = 81