Calc Optimization Help needed - should be easy question

I just don't know if I am drawing the question out properly.

Question: A fence is 1.5m high and is 1m from a wall. A ladder must start from the ground, touch the top of the fence, and rest somewhere on the wall. Calculate the minimum length of the ladder.

The answer in the textbook is 4.5 m.


Here's what I thought the scene would look like. A ladder is leaned against the fence and stretches out even more to hit the wall. So I broke up the hypotenuse into 2 sections. One section can deal with the height of the fence and the angle its leaning at. The second part is the distance from the fence to the wall and it should be at the same angle.

So like, h1 = 1.5/ sin x

h2 = 1/cos x

So then getting the full length you would add up the two sections, find the derivative, and find the angle that gives the min. length. Then sub that angle back into the original function to find the length of the ladder.

I did this 2 times but never got it. So I'm guessing my idea is incorrect. Any ideas?

your idea was right !
either the figures in the q or the ans appear to be wrong !

one line ans: ( ( 1.5^(2/3) + 1^(2/3) )^(3/2) = 3.51 m
▬▬▬▬▬▬

derivation:
▬▬▬▬

........../|
.....y./..|
L..../θ..|
..../|.....|
x/..|.h..|
/θ..|.....|
...... d

let θ be the angle the ladder makes with the ground
& L ft be the length of ladder = x+y
= (h/sinθ) + (d/cosθ)
dL/dt = h cosθ/sin^2 θ - dsinθ/cos^2 θ = 0 for minima
putting under common denominator, numerator
hcos^3 θ - dsin^3 θ = 0

tan^3 θ = h/d
tanθ = (h/d)^(1/3)

we can thus visualise a compressed rt. angle triangle
with base = d^1/3, ht. = h^1/3 and (hence) hypotenuse = L^1/3

applying pythagoras' theorem, L^(2/3) = h^(2/3) + d^(2/3)
L = ( (h^(2/3) + d^(2/3) )^(3/2) , i.e.
L = ( (1.5^(2/3) + 1^(2/3) )^(3/2)

L = 1.5/sinA + 1/cosA
dL/dA = -1.5cosA /sin^2A + sinA/cos^2A= 0 for minimum
-1.5cosA +sinA*tan^2A = 0
tan^3A = 1.5
A = 48.86 degrees
L = 1.5/sin48.86 + 1/cos48.86 = 3.51 m