Calculus Optimization Question?

Question

An eight-foot fence that stands on level ground is one foot from a telephone pole. Find the shortest ladder that will reach over the fence to the pole.

This is an extra credit question for my BC Calc class. I'm not sure how to start this

Answer 1

Let the distance from the bottom of the ladder to the fence be d. The ladder lies across the top of a fence, making a right triangle with legs d, and 8ft. This triangle is similar to the triangle made by the ladder with legs d+1ft and x, the height up the pole. Since the triangles are similar 8ft/d = x/(d+1ft) or x = [(d+1ft)/d](8ft). Using this knowledge we know the square of the length is

s =([(d+1ft)/d](8ft))^2 + (d+1ft)^2

Since the length is minimized when the square of the length is minimized, just take the derivative of the above ds/dd = 0 to find the distance from the base of the ladder to the fence, d, that minimizes the length.

Answer 2

assume the base of the ladder is x feet beyond the fence. Let the length of the ladder be L.
L is the hypoteneus of a right triangle whose legs are x+1 (along the ground) & h, the height at which the ladder meets the pole.
But by similar triangles, h/(x+1) = 8/x
hx = 8x+8
h = 8 + 8/x
L^2 = (x+1)^2 + (8+8/x)^2

now, find dL/dx and set to 0. solve for x and calculate L