A two-pen corral is to be built. The outline of the corral forms two identical adjoining rectang

A two-pen corral is to be built. The outline of the corral...

Key Concepts

Area Maximization in Geometric Figures

In problems involving optimizing area, geometrical relationships are combined with algebraic methods to determine the dimensions that yield the maximum area. This involves setting up an area function based on the given shape, substituting any constraints, and then using optimization techniques to find the best dimensions under the given conditions.

Using Constraint Equations to Eliminate Variables

This technique involves using the relationships provided by constraints (such as a fixed total length or perimeter) to represent one variable in terms of another. By reducing the number of independent variables, the optimization problem is simplified into a single-variable problem, making it easier to analyze and solve.

Quadratic Function Optimization

Many optimization problems reduce to finding the extremum of a quadratic function. Since quadratic functions have a parabolic shape, their maximum or minimum value occurs at the vertex. This can be determined algebraically by converting the quadratic expression into vertex form or using the formula -b/(2a) from the standard form.

Optimization Under Constraints

This concept involves finding the maximum or minimum value of a function (often related to area, volume, cost, etc.) subject to a limitation or fixed resource, such as a total amount of materials or perimeter. It requires expressing one or more variables in terms of the others using the constraint, enabling the remaining expression to be optimized.

Transcript

00:01 So i'm looking at a figure that is two side -by -side rectangles, so like this shape.

00:09 And i'm told that i have 120 feet for the perimeter, and i want to maximize the area.

00:24 So my primary equation, my focus of this problem is the area, and that's what i want to go through and find critical numbers for.

00:31 And then my secondary equation to use is that perimeter.

00:35 So based on this shape given, i'm going to go ahead and say that i have three widths from the left, the right, in the middle.

00:49 But i am just going to say i have two lengths looking at the entire base of the figure.

00:55 So for the perimeter, which is 120, i'd have to add up all those sides.

00:59 So it's three ws and two l's.

01:02 And then for the area, which is my primary focus, i would still just do length times width.

01:12 Because to fill in all this area, i just have to multiply that outside rectangle length times width.

01:22 So in order to find the critical numbers for area, i'm going to need to solve for just one of the variables.

01:28 So i can isolate either one, l or w.

01:31 And it doesn't matter.

01:32 I think i'm just going to solve for l just because.

01:35 So i would subtract over the 3w and divide by 2.

01:39 So that gets l by itself.

01:41 And this allows me to fill in to the formula for area.

01:44 So it's 120 over 2 minus 3w over 2 as the whole length multiplied times width.

01:53 Before i take its derivative, i would rewrite this as 60w simplifying that fraction minus three halves w squared.

02:03 So i distributed and simplified a little bit there.

02:06 So now when i take that first derivative to find the critical numbers, i'm going to write 60 minus 3w.

02:13 Power rule for each term...

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