Maxima and Minima Use the formulas of this section to give an alternative solution to the indica

Name: Problem 60, Chapter 3: Algebra and Trigonometry
Uploaded: 2020-07-17T17:38:11-08:00
Duration: 5 min
Channel: Cinsy Krehbiel
Description: Maxima and Minima Use the formulas of this section to give an alternative solution to the indicated problem in Focus on Modeling: Modeling with Functions on pages 237-244 . Problem 22

step 1 So we're going to start with fencing. We want to enclose the shape that's drawn here. The total

Maxima and Minima Use the formulas of this section to give...

Key Concepts

Optimization

Optimization involves finding the maximum or minimum values of a function under given conditions. In calculus, this typically means using derivatives to determine where a function attains its highest or lowest values within a certain domain, which has applications in many fields such as economics, engineering, and the natural sciences.

Critical Points

Critical points are points in the domain of a function where its derivative is zero or undefined. These points are potential candidates for local maximum or minimum values and serve as the starting point for determining the function’s optimal behavior.

First Derivative Test

The first derivative test involves analyzing the sign changes of the derivative around the critical points to determine whether a point corresponds to a local maximum, local minimum, or neither. This method is crucial in understanding the behavior of functions and identifying their extreme values.

Second Derivative Test

The second derivative test uses the second derivative of a function to determine the concavity at the critical points. A positive second derivative at a critical point indicates that the function is concave up (suggesting a local minimum), while a negative value indicates concavity down (suggesting a local maximum), providing a useful check on the first derivative test.

Transcript

00:02 So we're going to start with fencing.

00:05 We want to enclose the shape that's drawn here.

00:08 The total amount of fencing we have is 750 feet.

00:13 And we want to make a rectangular enclosure that's separated into four parts.

00:18 So we're going to have one, two, three, four, five times x, and we're going to have two times w.

00:30 It's going to be our total fencing.

00:33 So 750.

00:34 We'll put the units back in when we're done.

00:36 We want to maximize the area.

00:40 So, well, what's the area? the area is going to be x times w.

00:46 And to maximize that, we're going to want to make it a function of just one variable.

00:50 So let's make it a function of x, which means we're going to have to substitute something in for the w that's in terms of x.

00:58 So if we take our fencing formula, we're going to say that 2w equals 750 minus 5x.

01:10 And if we divide both sides by two, we get that w is equal to 375 minus 2 .5x.

01:23 So now we can substitute that in for our w and we'll have a function in x.

01:31 So 375 minus 2 .5x.

01:39 So let's go ahead and distribute the x, and that's going to be 375x minus 2 .5x squared.

01:50 That's a quadratic.

01:51 We'll rearrange it so we don't get lost in the variables.

01:55 So negative 2 .5x squared plus 375x.

02:04 And there's no constant.

02:05 So that's our area.

02:06 And that was the first part of the problem was to make a function or an equation for the area of the enclosure.

02:16 The second part is to find the dimensions that would maximize the area, to find the maximum area...

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