Solve Fence & Building Problem: Get Optimization Help

In summary, the shortest ladder that will reach the wall of the building from the ground is 16.65 ft.
  • #1
Jay9313
40
0
So I have a problem. The problem says:
A fence 8 ft tall rubs parallel to a tall building at a distance of 4 ft from the building. What is the length of the shortest ladder that will reach from the ground over the fence to the wall of the building?

I drew a picture to help, but I can't draw it on here. I drew a right triangle, it's fairly large. Halfway through the base, there's a straight line up representing the fence. I then made the hypotenuse of the entire triangle L. The wall is h.


1.I then set up a proportion. It's (x/8)=((x+4)/(h))
2.I solved the proportion for x and got 32/(h-8)
3.I then said (x+4)^2+h^2=L^2
I plugged in the x value i solved the proportion for into the above equation and got
(((4h)/(h-8))^2)+(h^2)^(1/2))=L (It might help to right this out instead of looking at it.

Now I have to take the derivative of this and set it equal to zero.. But I keep messing it up.. Can somebody show me how to do it?
I have been working and realized I can take the derivative of the Pythagorean theorem mentioned in step 3. But I still have trouble solving this. can somebody please help?
 
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  • #2
This problem is a lot easier if you use trig functions. Start with 4+x = L*cos(θ) and write an expression for x in terms of θ using the small triangle. Substitute and differentiate L with respect to θ. It works out quite nicely.
 
  • #3
Jay9313 said:
1.I then set up a proportion. It's (x/8)=((x+4)/(h))
2.I solved the proportion for x and got 32/(h-8)
3.I then said (x+4)^2+h^2=L^2
I plugged in the x value i solved the proportion for into the above equation and got
(((4h)/(h-8))^2)+(h^2)^(1/2))=L (It might help to right this out instead of looking at it.

Now I have to take the derivative of this and set it equal to zero.. But I keep messing it up.. Can somebody show me how to do it?

I just did this problem with a student this past week (Stewart, Section 4.7, no?). The trick if you do this algebraically is not to simplify your expression too soon. It will also be better if you eliminate h , rather than x . Your Pythagorean result becomes

[tex]L^{2} = (x + 4)^{2} + (\frac{8 \cdot (x+4)}{x})^{2} . [/tex]

There are two things that will be helpful to do. One is to factor out the ( x + 4 )2. The other is to differentiate the expression for L2 implicitly with respect to x . When you are looking to minimize a length given by the Pythagorean Theorem or the distance formula, since that length is a positive number, minimizing L2 also minimizes L . You avoid dealing with a radical until you absolutely have to (if you need to at all).

We will then have

[tex] \frac{d}{dx} L^{2} = \frac{d}{dx} [ (x + 4)^{2} [ 1 + \frac{64}{x^{2}}] ] = 0 [/tex]

[tex]\Rightarrow 2L \cdot \frac{dL}{dx} = 2 \cdot (x + 4) ( 1 + \frac{64}{x^{2}}) + [ (x + 4)^{2} ( - \frac{128}{x^{3}}) ] = 0 [/tex]

Since L is not equal to zero , we can just focus on getting the right-hand side of this equation to be zero. Also, x ≠ -4 , so we can divide that factor out of the equation; we can also take out a factor of 2 . This will leave

[tex] ( 1 + \frac{64}{x^{2}}) + [ (x + 4) ( - \frac{64}{x^{3}}) ] = 0 \Rightarrow 1 + \frac{64}{x^{2}} = (x + 4) ( \frac{64}{x^{3}}) \Rightarrow 1 + \frac{64}{x^{2}} = \frac{64}{x^{2}} + \frac{256}{x^{3}} \Rightarrow \frac{256}{x^{3}} = 1 . [/tex]

You can take it from there (I'll tell you that the minimized length is ≈16.65 ft.)

The trigonometric method hotvette describes is the one the solver for Stewart's solution manual used. It is "easier" in the sense that the differentiation to find the critical value for [itex]\theta[/itex] is simpler; you pay for it later in extracting the minimal length for the ladder (particularly if you are solving for the "exact" value). You get a different but equivalent expression for the length, which is confirmed when you calculate the decimal approximation.
 

1. How can optimization help solve fence and building problems?

Optimization techniques use mathematical algorithms to find the most efficient solution to a problem. In the case of fence and building problems, optimization can help determine the most cost-effective and structurally sound design for a given space.

2. What factors are considered in optimizing fence and building designs?

Factors such as budget, materials, location, and intended use of the fence or building are all taken into account when using optimization techniques. Other variables, such as wind and seismic load, may also be considered for structural optimization.

3. Can optimization help with both small and large scale projects?

Yes, optimization techniques can be applied to projects of any scale. Whether it is a small backyard fence or a large commercial building, optimization can help find the optimal design based on the given parameters.

4. Is optimization a cost-effective solution?

While the initial cost of using optimization techniques may vary, it can ultimately save time and money in the long run by finding the most efficient design. An optimized design can also reduce maintenance and repair costs in the future.

5. Are there any limitations to using optimization for fence and building problems?

Optimization techniques rely on the input of accurate data and assumptions, so it is important to have a thorough understanding of the problem at hand. Additionally, some factors may be difficult to quantify or may not be included in the optimization process, so it is important to consider all aspects of the project.

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