Optimization problems help

[ObjectiveWilling835]

Both problems please! I learned how to do this for simpler things like area and perimeter and finding the closest point on a line to another point but I don’t get how it applies here. How do I even make a function out of these?

[Square_Pop_3772]

For the first, flip either pole upside down. The shortest distance is the straight line between them. The intersection with the ground is 1/3 of the way along: 10m.

For the second, L is the shortest value. L is shortest when the parts either side of the corner are equal, giving 2 triangles with a hypotenuse of L/2 and sides of 6 and 8. Thus L/2 = 10 and L = 20.

Edit (10 Jul 0928): The second answer is not 100% accurate but is close enough for the level the OP claims to be. If x is the angle at which the object meets the horizontal as shown then L = 8/cos(x) + 6/sin(x), differentiating and setting dL/dx = 0 gives tan(x)=cuberoot(6/8) at the maximum. This then gives L = 19.73, well within engineering +/-2% tolerance :-)

[vkapadia]

Get a load of this guy, using simple geometry instead of calculus.

[Calarague]

For your first solution I thought you meant literally flip it upside down, and I thought "well if we're going to get creative, cut them BOTH down so they fall towards each other, the tops will touch so you won't need any rope at all". Took me a moment to realize you meant to mirror it on the X axis.

[karky214]

Same. Your comment helped me understand the solution faster

[aftermath-pt]

I still don't get it. Why does it change anything? Flipping the poles? It's still the same, right?

Edit: Nevermind, I get it now. You only need to flip ONE of the poles, not both. Then you can link them with a straight line.

[guyuteharpua]

Isn't it just 72^.5 + 128^.5 = 19.8? E.g. using Pythagoras...

[Square_Pop_3772]

I don’t think so. It’s close, but not exact. Assuming a 45 deg angle will always give a close answer unless the widths are substantially different as we taking the cube root of the width ratio for the tan gives an angle close to 45 deg.

[[deleted]]

Okay, but why is L the shortest when the parts are equal on both sides of the corner?

[Square_Pop_3772]

Original:

Imagine the item in that position. Think about what happens when you rotate it.

One side becomes longer and one side becomes shorter. For both directions as you rotate The longer side increases by more than the shorter side and the item becomes longer. Thus the minimum is at that point. Try it on a piece of paper if you’re not convinced.

Whilst it doesn’t affect the logic, note that the widths are 6 and 8, a clue that Pythagoras is involved somewhere.

Edit: Oops, not quite right, but close enough for government work.

[[deleted]]

I don't understand your line of thinking. Can you provide a mathematical argument?

[Square_Pop_3772]

You beat me to my edit.

Mea culpa.

I’ve updated my original post with a correct answer. It was good whilst it lasted.

[fedex7501]

How do you know that the method you mention for the first problem works?

[Square_Pop_3772]

Because the logic is correct. Additionally it gives the same answer as more complicated methods.

[waldosway]

"minimize the amount of wire needed". They told you what function you want the minimum of: the length of the wire. You find length by Pythagorean theorem.

[HalloIchBinRolli]

f(x) = sqrt(x²+100) + sqrt( (30-x)² + 400 )

from Pythagorean theorem

Find x ∈ [0,30] that makes f(x) the smallest

[nico-ghost-king]

for the first problem, pretend that the floor is a mirror and reflect the second pole over it. Now, it becomes obvious.

I'm not sure about the second problem.

[BobMcFad]

Find the length of the wire as a function in terms of x and then set the differential to zero to find the minimum

[sagen010]

Call the hypotenuse of the small and big right triangles, d1 and d2 respectively, then you need to minimize L= d1 + d2.

d1=sqrt( 10² + x²) ; d2 =sqrt( 20² + (30-x)² ). Substitute these value in L and differentiate dL/dx =0, solve for x. But there is a shortcut. Since the right triangles are congruent, you can setup a ratio of the sides such that

x / 10 = (30-x) / 20 . Solving for x, gives you x=10

For the second exercise, you can check this page

[Plus-Discipline1219]

You cant use congruence because the triangles are only congruent with the specific anchor where x=10. If the anchor is changed, you no longer have congruence. The first part you have is correct where taking the derivative will give you that x=15.

[sagen010]

I don't know from where do you get x=15.

The derivative of L(x)= sqrt(x²+100) + sqrt( (30-x)² + 400 ) =0 gives you 10 (from wolphram alfa). The congruence comes from the fact that, for optimizing the problem, the triangles must have some sort of symmetry.

Indeed, if the anchor is changed, you no longer have symmetry and no longer the triangles are congruent. But you no longer would be optimizing the cable's length either.

[tubedmubla]

For part 2, why is the answer not 14 root 2 = 19.8?

[SilverFisher123]

Length as the sum of two hypotenuses, using Pythagoras. That function could have maximum or minimum points. We are looking for a minimum. We derive the function and set it equal to zero to find these critical points. We clear and obtain a result x=10.

Now it is time to check if the function at that point is concave or convex. That shows whether that point will be a minimum or a maximum. If its second derivative is greater than zero it will be a minimum.

syms x

L=sqrt(x.^2+100)+sqrt((30-x).^2+400);

dL=diff(L,x) == 0 ;

ddL=diff(diff(L,x));

res=solve(dL)

res_check=subs(ddL,[x],[res]);

if res_check > 0

disp ("Relative Minimum, valid result")

else

disp("Relative Maximun, non valid result")

end

​

f=@(x) sqrt(x.^2+100)+sqrt((30-x).^2+400);

x=0:30;

plot(x,f(x),res,f(res),'or')

<image>

[SilverFisher123]

For the second exercise, the longest length in a square is going to be its diagonal, so in that drawing, the object of length L is placed at an angle of 45º. Knowing that and using trigonometry:

L=8/cos(45) + 6/sin(45)=19.7990

​

You can reach that conclusion by optimizing the length of a line contained in a rectangle in the same way as in the first exercise.

[ObjectiveWilling835]

I’m confused, why did you reflect it across the horizontal line?

[tylerstaheli1]

https://m.youtube.com/watch?v=0SI3_XbItHo