How to solve this problem?

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1 Answer
Jan 6, 2018

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#