Posted by **lala** on Saturday, April 10, 2010 at 11:43pm.

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)

- Calculus -
**drwls**, Sunday, April 11, 2010 at 7:58am
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.

- Concur with dwrls, book is wrong - Calculus -
**Reiny**, Sunday, April 11, 2010 at 9:31am
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

- Calculus -
**Anonymous**, Monday, April 15, 2013 at 5:27pm
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

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