Revisit t the ladder optimization problem

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SUMMARY

The discussion focuses on solving the ladder optimization problem involving a fence and a house. The goal is to find the shortest ladder length that extends from the ground, over a 6 ft high fence, to a 20 ft high house, positioned 8 ft away from the fence. The derived formula for the minimum ladder length is \(L_{\min}=\left(d^{\frac{2}{3}}+h^{\frac{2}{3}} \right)^{\frac{3}{2}}\), with \(d=8\) ft and \(h=6\) ft, resulting in a minimum length of approximately 19.73 ft. The use of implicit differentiation and Lagrange multipliers is highlighted as key techniques in deriving the solution.

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karush
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This is a common homework problem but..

A fence $6$ ft high runs parallel to the wall of a house of a distance of $8$ ft
Find the length of the shortest ladder that extends from the ground,
over the fence, to the house of $20$ ft high
and the horizontal ground extends $25$ ft from the fence.

$$L=\sqrt{\left(x+8 \right)^2 +\left(20-y\right)^2 }$$

its assumed implicit derivative but really?

I graphed this and noticed the local min was about $19.7$

$$\sqrt{\left(x+8 \right)^2 +\left(\frac{48+6x}{x}\right)^2 }$$
 
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I used Lagrange multipliers to derive a formula here:

https://mathhelpboards.com/questions-other-sites-52/his-question-yahoo-answers-regarding-finding-shortest-ladder-will-reach-over-fence-9939.html

$$L_{\min}=\left(d^{\frac{2}{3}}+h^{\frac{2}{3}} \right)^{\frac{3}{2}}$$

Plugging in the given data:

$$d=8\text{ ft},\,h=6\text{ ft}$$

We find:

$$L_{\min}\approx19.73\text{ ft}$$
 

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