A corridor of width #a# meets a corridor of width #b# at right angles. Workmen wish to push a heavy beam on dollies around the corner, but they want to be sure it will be able to make the turn before starting. How long a beam will go around the corner ?

this is an optimization question that i don't understand. The answer is #l=(a^(2/3)+b^(2/3))^(3/2)#

1 Answer
George C.
Apr 6, 2018

#(a^(2/3)+b^(2/3))^(3/2)#

Explanation:

Let us locate the outer corner of the corridor at #(0, 0)# and the inner corner at #(a, b)#.

Consider lines of negative slope passing through #(a, b)# and the distance between their #x# and #y# intercepts.

The minimum distance then models the maximum feasible length of beam.

graph{((x-14)^100+(y-13)^100-10^100)((x-20)^100+(y-20)^100-20^100)(y+x/2-5)((x-4)^2+(y-3)^2-0.02)(x^2+(y-5)^2-0.02)((x-10)^2+y^2-0.02) = 0 [-5.625, 14.375, -3, 7]}

The equation of such a line can be written in point slope form as:

#y-b = m(x-a)#

The #x# intercept is then given by putting #y=0# to find:

#x = a-b/m#

and the #y# intercept by putting #x=0# to find:

#y = b-am#

The square of the distance between these intercepts is:

#(a-b/m)^2+(b-am)^2 = a^2-2ab 1/m + b^2 1/m^2 + b^2-2ab m + a^2 m^2#

The minimum will occur when the derivative of this with respect to #m# is zero. That is:

#0 = 2ab 1/m^2 - 2b^2 1/m^3-2ab+2a^2m#

Multiplying by #m^3/2# this becomes:

#0 = abm-b^2-abm^3+a^2m^4#

#color(white)(0) = a^2m^4-abm^3+abm-b^2#

#color(white)(0) = am^3(am-b)+b(am-b)#

#color(white)(0) = (am^3+b)(am-b)#

#color(white)(0) = (a^(1/3)m+b^(1/3))(a^(2/3)m^2-a^(1/3)b^(1/3)m+b^(2/3))(am-b)#

Note here that the last linear factor gives #m=b/a > 0#, so is not suitable.

Also the quadratic factor has only non-real solutions.

So we require #m = -b^(1/3)a^(-1/3)#

With this value of #m#, the square of the distance between the intercepts is:

#(a-b/m)^2+(b-am)^2 = (a+a^(1/3)b^(2/3))^2+(b+a^(2/3)b^(1/3))^2#

#color(white)((a-b/m)^2+(b-am)^2) = a^(2/3)(a^(2/3)+b^(2/3))^2+b^(2/3)(b^(2/3)+a^(2/3))^2#

#color(white)((a-b/m)^2+(b-am)^2) = (a^(2/3)+b^(2/3))^3#

So the distance is:

#sqrt((a^(2/3)+b^(2/3))^3) = (a^(2/3)+b^(2/3))^(3/2)#

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