A fence 8 ft tall runs parallel to a tall building at a distance of 4 ft from the building. What

A fence 8 ft tall runs parallel to a tall building at a...

Key Concepts

Similar Triangles

Similar triangles are used to establish proportional relationships between different segments of a geometric configuration. In these types of problems, when a ladder leans against a wall and passes over an obstacle like a fence, similar triangles can help relate the heights and distances of various sections of the setup, which is crucial for setting up the equations required for optimization.

Pythagorean Theorem

The Pythagorean Theorem is a fundamental tool in dealing with right triangles, relating the lengths of the sides. In ladder problems, it is frequently used to connect horizontal and vertical distances, allowing one to express the length of the ladder as the hypotenuse of a right triangle whose legs represent the distances on the ground and against the wall.

Reflection Method in Geometry

This method involves reflecting one of the constraints or endpoints across a boundary to convert a constrained path problem into an unconstrained one. By using reflection, a problem that initially involves an intermediate obstacle can be transformed into finding a straight-line distance between two points, thereby simplifying the process of determining the shortest path.

Geometric Optimization

Geometric optimization refers to the process of finding a shape or path that minimizes (or maximizes) a certain parameter under given geometric constraints. In problems involving ladders, fences, or walls, it often involves determining the optimal placement or length by analyzing how different geometric elements interact.

Transcript

00:01 For this problem, we are looking at a building with a fence being propped up next to it.

00:06 So let's start by drawing a picture.

00:09 It's always helpful to have a picture for this type of problem.

00:12 Even if you're not a great artist, i certainly am not.

00:15 But being able to see a problem really does help you solve it.

00:18 Okay, so here is our situation.

00:21 I have a building with a fence, and they are four feet apart.

00:25 And my fence is eight feet tall.

00:28 And i want to have a ladder that just clears that fence and i can go from the ground to the building.

00:37 And my question is, if i call that x, i want to minimize x, the smallest ladder that i can get away with.

00:47 So how are these different lengths and sizes related to each other? well, i'm going to introduce two more variables just so i can talk about the different pieces of this.

00:59 Y is the part from the fence down to the end of the ladder, and z is the height on the building up to where the ladder will reach.

01:09 Now, if i look at the entire big triangle, i can see with the pythagorean theorem, that y plus 4 squared plus z squared equals x squared.

01:21 Or if i take the square root of both sides, x is the square root of y plus 4 squared plus z squared.

01:29 So that's my x.

01:38 Unfortunately, i have two variables here.

01:40 I have y and z, and i need to get rid of one of them.

01:43 So i need to see how i can relate them to each other.

01:46 So let's look at the smaller triangles.

01:48 First, let's look at that small piece right here.

01:53 I'm just kind of shading in it a little bit.

01:55 Let's look at that one first.

01:57 If i take side on the right to the base, that ratio is 8 to y.

02:05 Now what if i look at the whole big triangle altogether? while the piece on the right is now z and the base is now y plus 4.

02:15 So i can set up a proportion.

02:21 I can cross multiply and then i can solve this for one of my variables here.

02:28 Let's just solve for y.

02:31 I get 32 equals yz minus 8y and i can factor out a y and get z minus 8.

02:39 So y is going to equal, and let's put this over here on the side, y equals 32 over z minus 8.

02:47 I can plug that into my equation for x.

02:53 So i have x is the square root of 32 over z minus 8.

03:00 And i need to be a little more careful.

03:01 I don't want these z and twos to look alike.

03:04 So 32 over z minus 8 plus 4 squared plus z squared.

03:15 Okay, here is my x.

03:17 If i want to find the minimum, i'm going to take the derivative, set it equal to zero, and see what i get.

03:26 Now before i do that, i'm just going to clean up this bit with the parentheses just a little bit.

03:32 So if i get rid of those parentheses, expand out that binomial.

03:37 What i get is 1 ,024 over z minus 8 squared.

03:49 Plus twice the outer and the inner, that gives me 256 over z minus 8 plus 16.

04:00 And i'm going to add z squared to it because that's there on the end.

04:05 Okay, now let's take the derivative.

04:08 I have a square root, so that means when i take that derivative, it's going to be one half, so that two goes on the bottom.

04:16 And this is everything under that radical.

04:23 Instead of copying the whole thing, i hope we can just accept that everything under that radical is there right now.

04:28 Now i have to multiply this by the derivative of what's under the radical sign.

04:34 So let's do this in pieces.

04:36 First, we have a quotient...

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