Calculus
posted by lala .
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 minimun lenght of the ladder.
I drew a diagram and i ended up with a right angle triangle. i made the hypotenuse the lenght of the ladder, h=1.5 and base=1. I then used the pytagorean theorem to solve for the hypotenuse, but my answer doesnt match the one at the back of the book(4.5m)

Let A be the angle that the ladder makes with the horizontal. The ladder's length L must be such that
1.5/sinA + 1/cos A = L
Compute dL/dA and set it equal to zero to get the angle A for which the length is minimum.
dL/dA = 1.5 cosA/sin^2A + sinA/cos^2A = 0
1.5 cos^3A = sin^3A
tan^3A = 1.5
tanA = 1.145
A = 48.86 degrees
L = 1.5/sin48.86 + 1/cos48.86
= 1.992 + 1.520 = 3.512
I don't agree with your book's answer, either. 
I tried it all algebraically, no trig.
let the ladder reach y m up the wall, and touch the ground x m from the fence.
So I had two similar right angled triangles and
1.5/x = y /(1+x)
xy = 1.5 + 1.5x
2xy = 3 + 3x
y = (3+3x)/(2x)
L^2 = y^2 + )1+x)^2
= [(3+3x)/(2x)]^2 + (1+x)^2
= (9 + 18x + 9x^2)/(4x^2) + 1 + 2x + x^2
=(9/4)x^2 + (9/2)x^1 + 9/4 + 1 + 2z + x^2
2L(dL/dx) = (18/4)x^3  (9/2)x^2 + 2 + 2x
= 0 for a max/min of L
(18/4)x^3  (9/2)x^2 + 2 + 2x
= 0
times 4x^3
18  18x + 8x^3 + 8x^4 = 0
8x^3(1+x)  18(1+x) = 0
(1+x)(8x^3  18) = 0
x = 1, silly answer or
x^3 = 18/8
x = cuberoot(18)/2 = 1.31037
sub that back into y = ..
y = 2.6447
then L^2 = y^2 + (1+x)^2
gave me
L = 3.5117 
dwls solution is corret but when you plug 48.86 degrees back into the original equation your calculator must be in rads and you will get 4.5m