Fencing a Horse Corral Carol has 2400 ft of fencing to fence in a rectangular horse corral. (a)

Fencing a Horse Corral Carol has 2400 ft of fencing to fence...

Fencing a Horse Corral Carol has 2400 $\mathrm{ft}$ of fencing to fence in a rectangular horse corral.
(a) Find a function that models the area of the corral in terms of the width $x$ of the corral.
(b) Find the dimensions of the rectangle that maximize the area of the corral.

Key Concepts

Quadratic Function Optimization

Many optimization problems, after substitution of the constraints, reduce to a quadratic function. The properties of quadratic functions, particularly the fact that they form parabolas, allow us to use methods like finding the vertex to determine the maximum or minimum values. The vertex of a parabola gives the optimal solution in cases where the quadratic represents the scenario being optimized.

Function Modeling

This concept involves representing a real-world situation with a mathematical function that expresses one variable in terms of another. In optimization problems, such as maximizing area or volume, establishing a relationship between different dimensions allows us to analyze how changing one variable affects the outcome, which is essential in determining optimal solutions.

Constraint Manipulation

Often, real-world problems include constraints that limit the values of the variables, such as a fixed perimeter or fixed amount of material. By using these constraints, one can express one variable in terms of another and substitute back into the equation for the quantity of interest, thus reducing the number of variables and simplifying the problem.

Transcript

00:01 Okay, so for this question, we are told that fencing a horse coral carol has 2 ,400 feet of fencing to fencing a rectangular horse coral.

00:11 And it says, find a function that models the area of the coral in terms of the width x of the coral.

00:16 Okay.

00:17 So we're going to call the area a, and it tells us that the width of the coral is x.

00:33 All right.

00:33 So for the moment, because it's going to be a rectangle, right? so one of the sides we can label x, and for the moment we can label the other one y.

00:44 But we don't want two different variables in our equation.

00:48 We just want it all to be one variable.

00:50 So we need to find a way to write y in terms of x.

00:54 And the way we're going to do this is that we know the perimeter.

00:57 So x plus x, because that's both these sides here, plus y, plus y, that's both the top and the bottom side, that's going to equal 2 ,400.

01:10 So we'll just collect terms here, 2x plus 2y equals 2 ,400.

01:16 So then 2 ,400 minus 2x equals 2y.

01:22 And we have a common factor of 2 here.

01:23 So y is going to equal 1 ,200 minus x.

01:28 So then we can replace this up here.

01:31 Just for a moment.

01:33 We can replace this up here instead of having a y.

01:37 Instead of having a y we can have a 1 ,200 minus x and we have an x up here as well all right so we haven't quite found an equation for the area to find an equation for the area we'll just times these two together because the area is just the length times the width so the area or a so we can say a x equals x times by 1 ,200 minus x.

02:14 So that's going to be 1200x minus x squared...

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