Question A10​-ft-tall fence runs parallel to a wall of a house at a distance of 26 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 and the horizontal ground have infinite extent. The length of the shortest ladder is ft. ​(Round the final answer to the nearest tenth as needed. Round all intermediate values to the nearest thousandth as​ needed.)

Q. salh:- In this question giuen that,Now Ih \triangle A B D,\left.z^{2}=y^{2}+(26+x)^{2}+i\right) (by Phytagoras Thearem)In \triangle A B D and D C D E,\begin{array}{l}\frac{y}{26+x}=\frac{10}{x} \quad(\because \triangle A B D \sim \triangle C D E) \\y=\frac{10(26+x)}{x}\end{array}Put these value in equation (i)\begin{aligned}f(x)=z^{2} & =\left[\frac{10(26+x)}{x}\right]^{2}+(26+x)^{2} \\& =(26+x)^{2}\left[\frac{100}{x^{2}}+1\right] \\f(x) & \Rightarrow \frac{(26+x)^{2}\left(100+x^{2}\right)}{x^{2}} \\\text { Now } f^{\prime}(x) & =\frac{x^{2} \frac{d}{d x}(26+x)^{2}\left(100+x^{2}\right)-(26+x)^{2}\left(100+x^{2}\right) \frac{d}{d x} x^{2}}{x^{4}} \\& \Rightarrow \frac{x^{2}\left[(26+x)^{2}(2 x)+\left(100+x^{2}\right) 2(26+x)\right]-(26+x)^{2}\left(100+x^{2}\right) \cdot 2 x}{x^{4}} \\& \Rightarrow \frac{2 x^{2}(26+x)\left[x(26+x)+\left(100+x^{2}\right)\right]-(26+x)^{2}\left(100+x^{2}\right) \cdot 2 x}{x^{4}} \\& \Rightarrow \frac{2 x(26+x)\left[x\left(2 x^{2}+26 x+100\right)-(26+x)\left(100+x^{2}\right)\right]}{x^{4} 3} \\& \Rightarrow \frac{2(26+x)\left[x^{3}-2600\right]}{x^{3}}\end{aligned}Put f^{\prime}(x)=0, far critical numbers.\begin{aligned}\Rightarrow & \frac{2(26+x)\left(0 x^{3}-2600\right)}{x^{3}}=0 \\\Rightarrow & 2(26+x)\left(x^{3}-2600\right)=0 \\& {[x=-26] \& \quad x=\sqrt[3]{\frac{2600}{10}} } \\& {[x=-26] \& \quad x=13.751 }\end{aligned}Again,\begin{array}{l}f^{\prime \prime}(13.751)=\frac{2\left[(13.751)^{4}+5200(13.751)+202800\right]}{(13.751)^{4}} \\\Rightarrow 17.343>0 \text { (minima) } \\\end{array}By second derivative test f(x) will be minimum at x=-26 and x=13.751. but Negetive value hot acceptatale. Hence x=13.75 hen minimum value. and where f(x) will the minimum z^{2} will bue =2415.8 \mathrm{ft}.Thergare, The length of the shortest lather 4 is 2415.8 \mathrm{ff}Ans. ...