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IPMAT Indore 2024 - QA (SA) Q5 Explanation
November 21, 2024
Explanation:
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…eqn.(i)
△AFE ~△ABC ( AA criteria)
∴ AF/AB=FE/BC
= 18−y/18 = x/8 ⇒ y=x−18…eqn.(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 Ans.