A fence 4 feet tall runs parallel to a tall building at a distance of 6 feet from the building. What is the length of the shortest ladder that will reach from the ground over the fence to the wall of the building?
Make a diagram
let the foot of the ladder be x ft from the fence
let the ladder reach y ft above the ground
I see similar triangle so set up a ratio
4/x = y/(x+6)
xy = 4x+24
y = (4x+24)/x
let the length of the ladder be L
L^2 = (x+6)^2 + y^2
= (x+6)^2 + [(4x+24)/x]^2
2L dL/dx = 2(x+6) + 2[(4x+24)/x] (x(4) - (4x+24))/x^2
= 2(x+6) - 8(x+6)(-24)/x^3
= 0 for a min of L
2(x+6) - 192(x+6)/x^3 = 0
times x^3
2x^3(x+6) - 192(x+6) = 0
2(x+6)(x^3 - 96) = 0
x = -6 , which makes no sense
or
x = 96^(1/3) , (which is the cuberoot of 96)
x = 4.57886
sub back into L^2 = 197.3
L = 14.05 m
To find the length of the shortest ladder that will reach from the ground over the fence to the wall of the building, we can use the Pythagorean theorem.
The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
In this case, the ladder, the fence, and the distance from the fence to the building form a right-angled triangle.
Let's consider the ladder as the hypotenuse, the distance from the fence to the building as one side, and the height of the fence as the other side.
By applying the Pythagorean theorem, we can derive the following equation:
ladder^2 = distance^2 + height^2
In this case, the distance between the fence and the building is 6 feet, and the height of the fence is 4 feet. Plugging these values into the equation, we get:
ladder^2 = 6^2 + 4^2
Simplifying the equation:
ladder^2 = 36 + 16
ladder^2 = 52
To find the length of the ladder, we take the square root of both sides of the equation:
ladder = √52
Calculating the square root of 52:
ladder ≈ 7.21 feet
Therefore, the length of the shortest ladder that will reach from the ground over the fence to the wall of the building is approximately 7.21 feet.