Perimeter =2r+rθ=20
θ=r20−2r
Now, Area A=2r2θ
⇒A=2r2(r20−2r)=10r−r2
⇒drdA=10−2r=0
⇒r=5
dr2d2A(r=5)=−2<0⇒r=5 is the point of maxima
Amax=10×5−25=25
JEE Main 2017 — Mathematics Calculus
Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is:
Held on 2 Apr 2017 · Verified 6 Jul 2026.
12.5
10
25
30
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