MHB Revisit t the ladder optimization problem

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The discussion focuses on finding the shortest ladder that reaches over a 6 ft high fence to a 20 ft high house, with the fence located 8 ft from the house. The formula for the ladder's length is derived using implicit differentiation and Lagrange multipliers, leading to a minimum length of approximately 19.73 ft. A graph of the function indicates a local minimum around this value. The calculations confirm that the optimal ladder length aligns with the derived formula. The problem illustrates a practical application of optimization techniques in calculus.
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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