To find the value for #x#, we must find the point on the graph where #T(x)# is minimal. To do so, we must differentiate #T(x)# and make it equal 0. #"Minimum point": T'(x)=0#
We know from the hint that #T(x)=sqrt(16+x^2)/20+(9-x)/55#
#T'(x)=d/(dx)[(16+x^2)^(1/2)/20+(9-x)/55]#
#color(white)(Xlllll)=d/(dx)[(16+x^2)^(1/2)/20]+d/(dx)[(9-x)/55]#
#color(white)(Xlllll)=(d/(dx)[(16+x^2)^(1/2)])/20+(d/(dx)[9-x])/55#
#color(white)(Xlllll)=(d/(dx)[16+x^2] * 1/2 * (16+x^2)^(1/2-1))/20+(d/(dx)[9]-d/(dx)[x])/55#
#color(white)(Xlllll)=((d/(dx)[16]+d/(dx)[x^2])( 1/2 * (16+x^2)^(-1/2)))/20+(0-1)/55#
#color(white)(Xlllll)=(0+2x * 1/2 * (16+x^2)^(-1/2))/20-1/55#
#color(white)(Xlllll)=(x(16+x^2)^(-1/2))/20-1/55#
#color(white)(Xlllll)=x/(20sqrt(16+x^2))-1/55#
Now we need to make #T'(x)=0#
#x/(20sqrt(16+x^2))-1/55=0#
#x/(20sqrt(16+x^2))=1/55#
#55x=20sqrt(16+x^2)#
#3025x^2=400(16+x^2)#
#3025x^2=6400+400x^2#
#2625x^2=6400#
#x^2=6400/2625=256/105#
#x=sqrt(256/105)=sqrt(256)/sqrt(105)=16/sqrt(105)~~1.56#