Given #x^2+4y^2=4#
#rArrcolor(white)("XXX")y = sqrt(1-x^2/4)#
Since the width of the rectangle is #2x# (from #-x# to #+x#)
and the width is #2y# (from #-y# to #+y#)
Area of rectangle:
#color(white)("XXX")= 2x xx 2y = 4xy#
#color(white)("XXX")=4*x * sqrt(1-x^2/4)#
#color(white)("XXX")=4* sqrt(x^2-x^4/4)#
This value will be a maximum when
#color(white)("XXX")f(x) = x^2-x^4/4# is a maximum
For #f(x)# to be a maximum,
#color(white)("XXX")#(df)/(dx) = 0#
#(df)/(dx) = 2x-x^3 = x(2-x^2)#
So for a maximum
either #x=0# (clearly extraneous) or #x^2=2#
If #x^2=2# (and #x>=0#)
then #x=sqrt(2)#
Substituting this back into #y = sqrt(1-x^2/4)#
gives #y = sqrt(2)/2#
Width #=2x = 2sqrt(2)#
Height #= 2y = sqrt(2)#
