A fence #8 ft# tall runs parallel to a tall building at a distance of #4 ft# from the building. What is the length of the shortest ladder that will reach from the ground over the fence to the wall of the building?

Let us set up the following variables:

# { (L, "Total length of ladder "AB, ft), (h, "Vertical height of ladder "OA, ft), (x, "Horizontal distance between ladder and fence "BC, ft) :} #

By Pythagoras, we have:

# \ \ AB^2 = OA^2 + OB^2 #
# :. L^2 = h^2 + (x+4)^2 # .... [A]

We also have similar triangles if we consider:

# triangle BCD# and #triangle OAB#

Which gives us the proportional relationship:

# OA:CD = OB:BC #
# :. h/8 = (4+x)/x => h = (8(4+x))/x #

Substituting this last expression for #h# into Eq [A] we get:

# L^2 = ((8(x+4))/x)^2 + (x+4)^2 #
# \ \ \ \ = ((8x+32)/x)^2 + (x+4)^2 #
# \ \ \ \ = (8+32/x)^2 + (x+4)^2 # ... [B]

Our aim is to now minimimze #L# wrt #x#, and this involves looking for critical points of the derivative, so differentiation wrt #x# we get:

# 2L (dL)/dx = 2(8+32/x)(-32/x^2) + 2(x+4) #
# \ \ \ \ \ \ \ \ \ \ = 16((x+4)/x)(-32/x^2) + 2(x+4) #
# \ \ \ \ \ \ \ \ \ \ = 2(x+4){1-256/x^3} #

At a critical point the derivative is zero and so we have the equation:

# 2(x+4){1-256/x^3} = 0 #

Leading to the solutions:

# x+4 = 0 => x= -4 #
# 1-256/x^3 = 0 => x = root(3)(256) #

We require that #x gt 0# so we discard the negative solution , leaving us with a single solution #x = root(3)(256)#

(We will omit the proof that this critical value corresponds to a minimum, we can verify this graphically, or use the second derivative test)

And so at this critical value, we can calculate the minimum ladder length using [B]:

# L^2 = (8+32/root(3)(256))^2 + (root(3)(256)+4)^2 #
# \ \ \ \ = 277.1476713 ... #

# :. L = 16.6477527 .... ft #