AP EAMCET Engineering 2018 Apr 22 Shift 1 Paper

Let, s=x2+y2
As the value of x+y=32 the expression will be
=x2+(32−x)2

ds

dx

=2x+2(32−x)(−1)
=2x−2(32−x)
=4x−64
For maxima and the minima substitute

ds

dx

=0 in above expression
4x−64=0
x=16
And the value of y is
y=32−x
=32−16
=16
Again after differentiation

d2x

dx

=4=+ive
As s is minimum when x=y=16
The minimum value of
s=162+162
s=256+256
s=512