A 12-ft-tall fence runs parallel to a wall of a house at a distance of 30 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 enter your response here

ft.

Community Answer

Honor CodeSolved 1 Answer

See More Answers for FREE

Enhance your learning with StudyX

Receive support from our dedicated community users and experts

See up to 20 answers per week for free

Experience reliable customer service

Get Started

Solution:-Now,\triangle A C E and \triangle B C D are similar\begin{aligned}& \frac{b}{30+b}=\frac{12}{a} \\\Rightarrow & a=\frac{12(30+b)}{b}\end{aligned}Now, apply pythagorus theorem in \triangle E C A\begin{aligned}\therefore l & =\sqrt{a^{2}+(30+b)^{2}} \\& \Rightarrow \sqrt{\left(\frac{12(30+b)}{b}\right)^{2}+(30+b)^{2}} \\& \Rightarrow \sqrt{\left(1+\frac{144}{b^{2}}\right)(30+b)^{2}} \\& \Rightarrow \frac{(30+b)}{b} \sqrt{144+b^{2}}\end{aligned}for extrema let l^{\prime}(b)=0\begin{array}{l}\Rightarrow \frac{30+b}{\not b} \times \frac{2 b}{\$ \sqrt{144+b^{2}}}+\sqrt{144+b^{2}} \times\left(\frac{-30}{b^{2}}\right)=0 \\\Rightarrow \frac{b^{2}(30+b)-\left(30 \sqrt{144+b^{2}} \times \sqrt{144+b^{2}}\right)}{b^{2} \times \sqrt{144+b^{2}}}=0 \\\Rightarrow \frac{30 b^{2}+b^{3}-30 b^{2}-4320}{b^{2} \sqrt{144+b^{2}}}=0 \\\Rightarrow b^{3}-4320=0\end{array}\begin{array}{l}\Rightarrow b^{3}=4320 . \\\Rightarrow b \approx 16.29\end{array}Then\begin{aligned}l & =\frac{30+(4320)^{1 / 3}}{(4320)^{1 / 3}} \sqrt{144+(4320)^{2 / 3}} \\& \approx 57.49 \mathrm{ft}\end{aligned}\therefore The length of the shortest ladder is 57.49 \mathrm{ft}. ...