How do you find the minimum of T = #sqrt(x^2+1)/3-(2-x)/5# (WITHOUT USING CALCULUS)?

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Answer 1 · 2018-02-27T17:47:39

See below.

#15T = 5sqrt(x^2+1)+3x-6# or

#15T+6 = P = 5sqrt(x^2+1)+3x#

This problem is equivalent to

Minimize #P = 5 y + 3 x#

subjected to

#y = sqrt(x^2+1)# or

#y^2 = x^2+1#

which is geometrically equivalent to:

Determine the line

#L-> 5y+3x=P# tangent to

#H-> y^2-x^2=1#

such that #P# is minimum.

Now if #L# is tangent to #H# then

#L nn H# should have only one solution.

So

#((P-3x)/5)^2-x^2=1# or

#P^2-6Px-16x^2-25=0# solving for #x# we have

#x = 1/16(-3P+ 5 sqrt(P^2-16))# and then

#P = pm 4# at #x = -(pm 3/4)#

so concluding we have

#x = -3/4# giving #P = 4# and consequently

#15T+6=4 rArr T = -2/15#

Attached a plot showing the tangency

enter image source here

NOTE

#y = sqrt(x^2+1)# is a one-leaf curve and
#y^2-x^2=1# has two leafs

The squaring procedure includes extraneous solutions...