A farmer wishes to use 400 yards of fencing to enclose a rectangular region and subdivide the re

A farmer wishes to use 400 yards of fencing to enclose a...

Key Concepts

System of Equations

This technique involves setting up and solving multiple equations simultaneously. In the context of the problem, one equation arises from the area constraint and another from the combined external and internal fencing lengths, which together determine the unknown dimensions.

Quadratic Equation Solution

Once the system of equations is reduced, a quadratic equation often emerges. The ability to solve such equations—through factoring or using the quadratic formula—is a key step in finding the specific dimensions that satisfy both the area and fencing constraints.

Area of a Rectangle

This concept involves calculating the area of a rectangle by multiplying its length by its width. It's fundamental in solving many geometry problems and sets up the equation that links the dimensions of the rectangle to the given total area.

Perimeter Constraint with Internal Divisions

In problems where a rectangle is subdivided, not only is the outer perimeter considered but also the additional fencing used for internal divisions. A clear understanding of how the internal segments contribute to the total length of fencing is essential to correctly form the constraint equation.

Transcript

00:01 So in this problem, we're given this diagram to represent the fencing that this farmer is going to use.

00:05 And we're told that he has a total of 400 yards of fencing that he's going to use to enclose a space with an area of 4 ,800 square yards.

00:14 Okay.

00:15 Well, because we're told that he has 400 yards of fencing, that's almost like finding the perimeter of all of our sides.

00:22 So to find the perimeter, we would have w, which is this length, but it happens four times.

00:27 So 4w would represent those distances in red, and this bottom side and the top side that are now in green are both l.

00:36 So we have plus 2l, and this was all equal to the total amount of fencing, which is 600.

00:41 Well, we're told that the total area enclosed is 15 ,000.

00:46 So almost pretend if these two sections of fencing weren't here, essentially you would just have a rectangle.

00:52 So to find the area of the rectangle, you multiply the length times the width.

00:55 So we know that l times w is equal to 4 ,800.

01:00 Okay, great.

01:02 So what we are going to do now is now we need to solve the system of equations.

01:07 So first thing i'm going to do is i'm going to divide both sides by w.

01:11 I'm going to isolate l.

01:13 So we have l equal to 4 ,800 divided by w.

01:17 So now we're going to go back to that first equation, which was 4w plus 2l is equal to 400.

01:25 Okay.

01:26 Okay, and i apologize.

01:28 I put 600.

01:29 This should have been 400.

01:30 I think i kept saying 400, but then didn't write it.

01:33 So this should have been 400.

01:35 Okay.

01:36 So what i'm going to do now, because each turns divisible by two, i'm going to divide both sides of our equation by two, just to make these numbers a little smaller.

01:43 So we'll have 2w plus l is equal to 200.

01:47 So now we're going to substitute for the 800 over w in place of l.

01:51 So we'll have 2w plus 4 ,800, divided by w, equal to 200.

01:57 Okay.

01:58 So next, we have to get rid of our fraction.

02:00 So we're going to multiply both sides of our equation by w.

02:04 W times 2w is 2w squared.

02:07 W times 4800 over w will just be 4 ,800 because it will cancel the w.

02:12 And then we have w times 200, which is 200 w.

02:17 So now i see we have a quadratic equation.

02:20 So we're going to set this equal to 0 by subtracting 200 w from both sides.

02:25 So we'll have 2 w squared minus 200 w plus 4800...