a 10 - ft - tall fence runs parallel to the wall of a house at a distance of 8 ft. find the length of the…

a 10 - ft - tall fence runs parallel to the wall of a house at a distance of 8 ft. find the length of the shortest ladder that extends from the ground to the house without touching the fence. assume the vertical wall of the house is 25 ft high and the horizontal ground extends 20 ft from the fence. let l be the length of the ladder, x be the distance from the base of the ladder to the fence, d be the distance from the fence to the house and h be the distance from the ground to the point the ladder touches the house. what is the objective function, in terms of x? l = (type an expression.)

a 10 - ft - tall fence runs parallel to the wall of a house at a distance of 8 ft. find the length of the shortest ladder that extends from the ground to the house without touching the fence. assume the vertical wall of the house is 25 ft high and the horizontal ground extends 20 ft from the fence. let l be the length of the ladder, x be the distance from the base of the ladder to the fence, d be the distance from the fence to the house and h be the distance from the ground to the point the ladder touches the house. what is the objective function, in terms of x? l = (type an expression.)

Answer

Explanation:

Step1: Use similar - triangles property

Let the height on the house where the ladder touches be (h). By similar triangles, (\frac{10}{x}=\frac{h}{x + 8}), so (h=\frac{10(x + 8)}{x}).

Step2: Apply the Pythagorean theorem

The length of the ladder (L) is given by the Pythagorean theorem. The horizontal distance from the base of the ladder to the house is (x + 8) and the vertical distance is (h). So (L=\sqrt{(x + 8)^2+h^2}). Substitute (h=\frac{10(x + 8)}{x}) into the formula for (L): [ \begin{align*} L&=\sqrt{(x + 8)^2+\left(\frac{10(x + 8)}{x}\right)^2}\ &=(x + 8)\sqrt{1+\frac{100}{x^2}} \end{align*} ]

Answer:

((x + 8)\sqrt{1+\frac{100}{x^2}})