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Consider the collection of all curves of the form y=abx2 that pass through the point (2,1) where a and b are positive real numbers. If the minimum area of the region bounded by y=abx2 and the xaxis is A sq. units then the value of AN is

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Solution


y=abx2
When x=2,y=1
1=a4ba=1+4b

Now, Area A=2a/b0(abx2)dx
A=2[axbx33]a/b0
A=2[aabb3aabb]
A=4a3/23b
A=43(1+4b)3/2b (a=1+4b)
Differentiating w.r.t. b, we get
dAdb=43⎢ ⎢ ⎢ ⎢6b1/2(1+4b)1/2(1+4b)3/2121bb⎥ ⎥ ⎥ ⎥
dAdb=23[12b(1+4b)(1+4b)3/2b3/2]
dAdb=231+4bb3/2(12b14b)

For max/min area, dAdb=0
b=18; a=32
By first derivative test, A has b=18 as point of minima.
Amin=43(32)3/222
Amin=48 sq. units

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