A Tricky Visual Calculus Problem

Calculus Level 2

What is the approximate length of the longest ladder (in feet) you can carry horizontally around the corner of the corridor shown here?

A B AB is the ladder and the coordinate of R R is ( 8 , 6 ) (8,6) .


The answer is 19.731.

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Venkatesh K by Venkatesh K , 21, India

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6 solutions

Let θ = ∠ O A B , \theta = \angle OAB, where O ( 0 , 0 ) O(0,0) is the origin. (Note that we are looking for 0 < θ < π 2 . 0 \lt \theta \lt \frac{\pi}{2}. ) Then the length of the ladder is

L = ∣ A R ∣ + ∣ R B ∣ = 6 sin ⁡ ( θ ) + 8 cos ⁡ ( θ ) , L = |AR| + |RB| = \dfrac{6}{\sin(\theta)} + \dfrac{8}{\cos(\theta)}, (A).

Now differentiate with respect to θ \theta and set equal to 0 : 0:

d L d θ = 6 cos ⁡ ( θ ) sin ⁡ 2 ( θ ) − 8 sin ⁡ ( θ ) cos ⁡ 2 ( θ ) = 6 cos ⁡ 3 ( θ ) − 8 sin ⁡ 3 ( θ ) sin ⁡ 2 ( θ ) cos ⁡ 2 ( θ ) = 0 \dfrac{dL}{d\theta} = \dfrac{6\cos(\theta)}{\sin^{2}(\theta)} - \dfrac{8\sin(\theta)}{\cos^{2}(\theta)} = \dfrac{6\cos^{3}(\theta) - 8\sin^{3}(\theta)}{\sin^{2}(\theta)\cos^{2}(\theta)} = 0

when 6 cos ⁡ 3 ( θ ) = 8 sin ⁡ 3 ( θ ) ⟹ tan ⁡ ( θ ) = 3 4 3 6\cos^{3}(\theta) = 8\sin^{3}(\theta) \Longrightarrow \tan(\theta) = \sqrt[3]{\dfrac{3}{4}}

⟹ θ = 0.737524472.... \Longrightarrow \theta = 0.737524472.... radians.

Now L → ∞ L \rightarrow \infty as θ → 0 \theta \rightarrow 0 and θ → π 2 , \theta \rightarrow \frac{\pi}{2}, so the critical point obtained must yield a minimum, which in turn corresponds to the longest possible ladder that can be carried around the corner of the corridor. Plugging the value for θ \theta found above into equation (A), we find that the desired length of the ladder is

L = 19.731 L = \boxed{19.731} feet to 3 decimal places.

17 Helpful 0 Interesting 0 Brilliant 0 Confused

sir , i have written a solution for this problem, tell me ,it is correct or not.

manish kumar singh - 5 years, 7 months ago

Brian sir, I have a doubt. The longest ladder can be of infinite length. The question should ask the length of shortest ladder. Am I wrong anywhere? Please correct me.

Rinkon Saha - 5 years, 1 month ago

for the length of the ladder to be maximum ,the ladder must pass through R(8,6). let A has coordinate (a,0) and B has coordinate (0,b). then eqn of straight line AB is x/a +x/b =1. this st. line passes through point R, HENCE we have, 8/a + 6/b =1. which gives b=(6 a)/(a - 8). now length of ladder is given by, L= sqrt(a^2 + b^2). PUT b in terms of a, we have L^2= a^2 +[ 36 a^2/(a-8)^2]. for length to be maximum d(L)/d (a) =0. which gives a =14.6038, and b= 13.2686. so longest length = sqrt(a^2+b^2)= 19.7313 (approx).

2 Helpful 0 Interesting 0 Brilliant 0 Confused

This is a great approach, but I think that you may have made a mistake after setting d L d a = 0 , \dfrac{dL}{da} = 0, since I'm getting different values for a a and b . b. Rewriting your equation for L 2 L^{2} as

L 2 = a 2 ( 1 + 36 ( a − 8 ) 2 ) L^{2} = a^{2}\left(1 + \dfrac{36}{(a - 8)^{2}}\right)

and then differentiating with respect to a a using the product rule, we find that

2 L d L d a = 2 a ( 1 + 36 ( a − 8 ) 2 ) + a 2 ( − 2 ∗ 36 ( a − 8 ) 3 ) 2L\dfrac{dL}{da} = 2a\left(1 + \dfrac{36}{(a - 8)^{2}}\right) + a^{2}\left(\dfrac{-2*36}{(a - 8)^{3}}\right)

⟹ d L d a = a L ( 1 + 36 ( a − 8 ) 2 − 36 a ( a − 8 ) 3 ) = \Longrightarrow \dfrac{dL}{da} = \dfrac{a}{L}\left(1 + \dfrac{36}{(a - 8)^{2}} - \dfrac{36a}{(a - 8)^{3}}\right) =

a L ( 1 + 36 ( a − 8 ) 2 ( 1 − a ( a − 8 ) ) ) = a L ( 1 − 36 ∗ 8 ( a − 8 ) 3 ) . \dfrac{a}{L}\left(1 + \dfrac{36}{(a - 8)^{2}}\left(1 - \dfrac{a}{(a - 8)}\right)\right) = \dfrac{a}{L}\left(1 - \dfrac{36*8}{(a - 8)^{3}}\right).

Then since a , L > 0 a,L \gt 0 we see that d L d a = 0 \dfrac{dL}{da} = 0 when

1 = 36 ∗ 8 ( a − 8 ) 3 ⟹ a − 8 = ( 8 ∗ 36 ) 1 3 = 2 ∗ 6 2 3 ⟹ a = 8 + 2 ∗ 6 2 3 = 14.604 1 = \dfrac{36*8}{(a - 8)^{3}} \Longrightarrow a - 8 = (8*36)^{\frac{1}{3}} = 2*6^{\frac{2}{3}} \Longrightarrow a = 8 + 2*6^{\frac{2}{3}} = 14.604

to 3 decimal places, in which case b = 6 a a − 8 = 13.268 b = \dfrac{6a}{a - 8} = 13.268 to 3 decimal places, and so L = 14.60 4 2 + 13.26 8 2 = 19.731 L = \sqrt{14.604^{2} + 13.268^{2}} = 19.731 to 3 decimal places, in agreement with my solution. Otherwise, good work. :)

(There are also a couple of typos, but these don't affect your equation for L 2 . L^{2}. The coordinates for B B should be ( 0 , b ) , (0,b), and the equation for the line A B AB should be x / a + y / b = 1. x/a + y/b = 1. )

Brian Charlesworth - 5 years, 7 months ago

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sorry sir, yes B is (0,b). i also differentiated L^2 with respect to 'a'. actually i am not good at Latex, so i directly wrote values after calculating on paper, and you are right ,i have done wrong calculation . a= 14.6038 and b= 13.268, so L =19.731.i have edited. Thank you.

manish kumar singh - 5 years, 7 months ago
Care No More
Aug 28, 2015

simple,imprecise solution "20"

2 Helpful 0 Interesting 0 Brilliant 0 Confused

i think u said it using the property of circumvented.but u can't tell that's the centre

Ridhu Paran - 5 years, 1 month ago

I did it in a similar way but it's not correct to consider OR = RA

Apoorva Singal - 11 months, 3 weeks ago
Sourabh Jain
Aug 16, 2015

just scale that and prove it approx value is 20 :P

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Hadia Qadir
Aug 12, 2015

The approximate length of the longest ladder (infeet) one can carry horizontally around the corner of the corridor shown is [8^(2/3 + 6^2/3]^3/2 = 19.731 ft.

1 Helpful 0 Interesting 0 Brilliant 1 Confused 
1
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answer**(2/3) = 8**(2/3) + 6**(2/3)

answer = 19.7313

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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