A rancher has 200 feet of fencing to enclose two adjacent...

A rancher has 200 feet of fencing to enclose two adjacent rectangular corrals (see figure). (FIGURE CAN'T COPY) (a) Write the area $A$ of the corrals as a function of $x .$ (b) Construct a table showing possible values of $x$ and the corresponding areas of the corral. Use the table to estimate the dimensions that will produce the maximum enclosed area. (c) Use a graphing utility to graph the area function. Use the graph to approximate the dimensions that will produce the maximum enclosed area. (d) Write the area function in standard form to find analytically the dimensions that will produce the maximum area. (e) Compare your results from parts (b), (c), and (d).

A rancher has 200 feet of fencing to enclose two adjacent rectangular corrals (see figure).
(FIGURE CAN'T COPY)
(a) Write the area $A$ of the corrals as a function of $x .$
(b) Construct a table showing possible values of $x$ and the corresponding areas of the corral. Use the table to estimate the dimensions that will produce the maximum enclosed area.
(c) Use a graphing utility to graph the area function. Use the graph to approximate the dimensions that will produce the maximum enclosed area.
(d) Write the area function in standard form to find analytically the dimensions that will produce the maximum area.
(e) Compare your results from parts (b), (c), and (d).

Key Concepts

Use of Value Tables

The creation of a table that lists different values of the variable and their corresponding function outputs helps in estimating the optimal solution. By observing trends and identifying the highest computed area, one can approximate the dimensions that maximize the area.

Graphical Analysis

Graphical analysis involves using a graphing utility to plot the function and visually estimate key features, such as the maximum point. This approach provides an intuitive understanding of the behavior of the quadratic function and aids in identifying optimal dimensions.

Constraints in Mathematical Modeling

Constraints are conditions or limitations, such as the fixed amount of fencing available, that must be incorporated into the model. These constraints allow us to reduce the number of independent variables and derive a function that accurately represents the scenario.

Quadratic Functions

A quadratic function is a polynomial of degree two that can be written in standard form. Its graph is a parabola, which either opens upwards or downwards. The vertex of the parabola represents the maximum or minimum value, making it a crucial tool for finding optimal solutions, such as maximum area in enclosing problems.

Function Modeling

This concept involves translating a real-world situation into a mathematical function that describes the relationship between variables. In this case, the area of the corrals is expressed as a function of one of its dimensions, linking geometry with algebra through an equation that models the area based on a chosen variable.

Optimization

Optimization refers to the process of finding the best solution under given constraints. In this context, it means determining the dimensions that yield the largest possible area for the corrals, which is achieved by finding the maximum value of the quadratic function.

Transcript

00:01 Hello there.

00:02 So we are given this problem where we say that a rancher has 200 ft of fencing to enclose two adjacent rectangular corrales.

00:12 I've drawn them here.

00:15 So we know that we only have 200 ft offense to work with.

00:21 And in our problem that's going to relate to the total perimeter because we know if we have 200 ft offense to work with and our perimeter is going to be exactly equal to 200 ft.

00:33 And let's recall that for these two rectangles, the perimeter is going to be the some of the side links.

00:39 We have four sides whose side link is x.

00:43 And three sides whose side length is why.

00:48 And we know that this is going to be equal to 200 ft.

00:54 So we can write this equation and we can solve for y as a function of x.

01:00 This implies that 200 -4 x divided by three is why.

01:10 And that's going to be useful as we solve this problem.

01:14 So we'll put this in a box.

01:16 The first part of this problem asks asks us to write the area of the corrals as a function of x.

01:23 Now, since there are two corrals and they're both rectangular, we know that the area is going to be x.

01:31 Y.

01:32 For the first corral plus xy.

01:36 Which will be two x.

01:37 Y.

01:38 And since we already have an equation for x or for y as a function of x.

01:44 We can say two x.

01:46 Y.

01:47 This equal 22 x times 200 -4 x divided by three.

01:57 Which we can take the three out which is going to be 2/3 x times 200 -4 x.

02:04 And notice that both 204 x are divisible by four.

02:07 And so once again we can factor out of four and we'll get 8/3 x times 50.

02:15 That's 200 divided by four minus x.

02:20 And so this expresses area has a function of x.

02:27 The second part of this problem be is asking us to create a table showing possible values of x and the corresponding areas of the corral.

02:40 So i'm going to make my table x aux.

02:46 And by looking at this equation, i can tell that when i plug in zero, i'm going to get zero and so i'm going to start there.

02:55 I'm going to start with zero in an area of zero and then i'm going to move up by five.

03:00 When i plug in five i'm going to get 600.

03:04 When i plug in 10 i'm going to get 1066 0.6, repeating make my table even longer and i plug in 15.

03:14 We'll get 1400 and i played in 20 1600 when i plug in 25 16 66 0.6, repeating 30 1600.

03:29 And now i've become a little suspicious because i know this is a quadratic equation.

03:35 Right? i know that it has a negative leading coefficient.

03:38 So the graph is going to look something like this and upside down parabola a parabola that opens down.

03:46 I know that parabolas are symmetric about their vertex and that the vertex for this parabola represents the maximum area because it is the maximum on the graph of the area.

04:02 Mm and so when i see this 1600 i think oh that's the metric about a point.

04:07 So let's see if we continue to see the symmetry...

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