The length of the shortest ladder that will reach from the ground over the fence to the wall of the building is 16.65 ft(rounding up to 2 decimal places)
The following diagram is followed to answer the question.
The following variables are to be defined, in accordance with the diagram.
L is the total length of the ladder whose height is to be assumed as AB ft,
h is the vertical height of the ladder, assumed to be OA ft
x is the horizontal distance between the ladder and fence, assumed BC ft.
According to Pythagora's theorem :
[tex]AB^{2} =OA^{2} +OB^{2}[/tex]
∴[tex]L^{2} =h^{2} +(x+4)^{2}[/tex]........(i)
It is to be noticed that the triangles △BCD and △OAB are similar.
This gives rise to the proportional relationship:
OA:CD=OB:BC
∴ [tex]\frac{h}{8} =\frac{(4+x)}{8} \\ = > h=\frac{8(4+x)}{8}[/tex]
Substitution of h in expression (i) gives the following result:
[tex]L^{2} =(\frac{8(4+x)}{x} )^{2} +(x+4)^{2} \\= > L^{2} = (\frac{32+8x}{x})^{2} +(x+4)^{2} \\ = > L^{2}=(\frac{32}{x} +8)^{2}+(x+4)^{2}[/tex]..........(ii)
The intention is to minimise L w.r.t to x.To achieve this the critical points of the derivative has to be found.
Therefore differentiating the above equation w.r.t x,
[tex]2L\frac{dL}{dx} =2(8+\frac{32}{x} )(-\frac{32}{x^{2}} )+2(x+4)[/tex]
[tex]= 16(\frac{ x+4}{x})(-\frac{32}{x^{2} })+2(x+4)\\ =2(x+4)(1- \frac{256}{x^{3}})[/tex]
At critical points, the derivatives of any equation are 0. This gives us:
[tex]2(x+4)(1- \frac{256}{x^{3}})=0\\x+4=0 , x=-4 \\or,\\(1- \frac{256}{x^{3}})=0 = > x=\sqrt[3]{256}[/tex]
The distance cannot be a negative number. therefore x<0 values are to be rejected, giving the solution that, [tex]x=\sqrt[3]{256}[/tex].
Assuming the second order derivative criteria is fulfilled for the critical points, substituting the value of x in (ii) it is seen:
[tex]L^{2} =(8+\frac{32}{\sqrt[3]{256} } )+(\sqrt[3]{256} +4)^{2}\\L^{2}=277.1476...ft\\L=16.67....ft[/tex]
The answer is that the length of the shortest ladder that will reach from the ground over the fence to the wall of the building is 16.65 ft(rounding up to 2 decimal places)
Learn more about Pythogora's Theorem here:
brainly.com/question/343682
#SPJ4
