A gardener wishes to use 600 feet of fencing to enclose a rectangular region and subdivide the r

step 1 So in this problem, we have this diagram that I put from your book. It says, The Gardiner wishes

A gardener wishes to use 600 feet of fencing to enclose a...

Key Concepts

Quadratic Equations

Many geometric problems with constraints result in quadratic equations when one variable is substituted from one equation into another. Understanding how to form, manipulate, and solve quadratic equations is essential as it often leads to finding the optimal or required dimensions of the enclosure.

Systems of Equations

When multiple constraints are present—like a fixed amount of fencing (perimeter constraint) and a specified area—the problem naturally leads to forming a system of equations. Solving these equations simultaneously allows one to determine the unknown dimensions of the shape.

Perimeter and Area Relationships

This concept deals with understanding how the perimeter of a shape is determined by its dimensions, while the area is a product of those dimensions. In problems involving enclosures, it is crucial to set up the relationships that connect the amount of available boundary material (such as fencing) with the dimensions required to achieve a specific area.

Transcript

00:01 So in this problem, we have this diagram that i put from your book.

00:04 It says, the gardiner wishes to use 600 feet of fencing to enclose this rectangular region, and that the total area that's enclosed is 15 ,000 square feet.

00:14 Well, if we're told that he's going to use 600 feet of fencing, essentially that's the perimeter, but that would also include this length right here.

00:22 So essentially, we can think of this as the perimeter is equal to 600, and we're told that the area of the whole space is equal to 15 ,000.

00:29 So now we want to write an equation to represent both of these.

00:33 Well, the perimeter, that would be when we add up all the sides.

00:37 Well, in this case, each of these sides is w.

00:40 So we have 3w.

00:42 And then the other part of the fencing have both length l.

00:45 So plus 2l, and this would equal to the total amount of fencing, which is 600.

00:51 Now, to find the area, we actually don't have to worry about this section right here.

00:55 Essentially, we're finding the area of a rectangle, which is the length times the width.

00:59 And this is equal to, in this case, 15 ,000.

01:03 Okay, so now we have our two equations.

01:06 We just need to solve this system.

01:08 Well, i'm going to take the second equation and solve for l by dividing both sides by w.

01:13 So we'll have l equal to 15 ,000 divided by w.

01:18 So what we're going to do is we're going to substitute this expression in place of l in our first equation.

01:24 So we're going to have 3w plus 2 times 15 ,000 over w.

01:30 And this is equal to 600.

01:33 So the next thing i'm going to do is i'm going to distribute the 2.

01:36 So we're going to have 3w plus, well, 2 times 15 ,000 over w would be 30 ,000 over w.

01:44 And now to get rid of our fraction, we're going to multiply each side of our equation by w.

01:49 Well, w times 3w is 3w squared.

01:54 W times 30 ,000 over w is just 30 ,000 because the w would cancel.

01:59 And w times 600 is 600 w.

02:02 Okay.

02:04 So now i see we have a quadratic, so we need to set it equal to zero.

02:07 So we'll subtract each side by 600, or 600 w, i should say...

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