Answered By Mingrui (Mark) J

This is a classical question on pythagorean theorem, the ground and the fence and the ladder construct a right-angled triangle where the base and height is the distance from the wall to the fence (on the ground) and the height of the fence, the ladder is the hypotenuse of the triangle. 

we have base = 3 ft.; height = 8 ft.; hypotenuse = unknown.

from Pythagorean theorem,

   $hypotenuse^2=base^2+height^2=8^2+3^2=64+9=72ft^2$hypotenuse²=base²+height²=8²+3²=64+9=72ƒ t²  

   $hypotenuse=\sqrt{hypotenuse^2}=\sqrt{72}=6\cdot\sqrt{2}$hypotenuse=√hypotenuse²=√72=6·√2 

   $ladder=hypotenuse=6\cdot\sqrt{2}ft.=8.485ft.$ladder=hypotenuse=6·√2ƒ t.=8.485ƒ t.    

Answered By Rohtaz S

This question is about calculating the minimum length of the ladder.

Given:

The height of the fence is equal to 8ft and the distance between the fence and wall is 3=ft

Let us assume that the angle between the floor and ladder is X^o and the length of the ladder is Y ft

=> Y=3*secant(X)+8*Cosec(X)

The length of the ladder Y is a function of angle X and in order to calculate the minimum valus of Y with respect to X, we have to differentiate the equation of Y above with respect to X.

After solving the equation, we get dY/dX = tan³X-(8/3)

To get the minimum, solve for tan³X-(8/3)=0

=> X=tan-1((8/3)^1/3)

Solve to get X=54.2^o

So, the length of ladder will be minimum when X=54.2^o

Minimum value of Y =  Y=3*secant(54.2)+8*Cosec(54.2) ~ 14.99ft

Verify that the length is a minima by double derivative of Y with respect to X

d²Y/dX² = 3 sec^2(X) tan^2(X)

Using X = 54.2, we get the double derivative positve (6.277) which is a proof that the point is a minima.

3 sec^2(54.2) tan^2(54.2) = 6.277

The answer can also be verified by puttiing in other random values of X between 0^o to 90^o angle to get

Y > 14.99 ft