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

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

Key Concepts

Modeling with Functions

This concept refers to the process of representing real-world situations with mathematical functions. By using functions to simulate scenarios, one can apply calculus techniques to systematically analyze and solve practical problems.

Optimization

Optimization is the process of finding the best solution, such as maximizing profit or minimizing cost, subject to certain conditions. It often involves applying calculus methods like differentiation to model and solve real-world problems.

Second Derivative Test

The second derivative test is a tool used to classify critical points by examining the concavity of the function. A positive second derivative at a critical point indicates a local minimum, while a negative second derivative suggests a local maximum.

Critical Points

Critical points occur where the derivative of a function is zero or does not exist. These points are crucial in optimization problems because they are the most likely candidates for local maximum or minimum values.

Derivatives

Derivatives are used to calculate the rate of change of a function. In the context of finding extrema, derivatives help in identifying where the function's slope becomes zero or undefined, which are potential points for maximum or minimum values.

Maximum and Minimum Values

This concept involves determining the largest and smallest values a function can attain within a given domain. It is central to calculus and optimization, particularly when analyzing the behavior of functions to identify optimal outcomes in various applications.

Transcript

00:02 So we want to build a window and it has to have a perimeter of 30 feet.

00:12 And we're trying to build it such that it has the maximum amount of light, which is really saying the maximum area.

00:21 So this window is made up of a rectangle with the dimensions h by x and a semicircle.

00:29 So that's exactly half of a circle.

00:31 So that circle has a radius of r.

00:36 And that's exactly half of x.

00:38 So we can write that right here, r equals x over two.

00:43 So the perimeter is going to be x plus h plus the perimeter of half of that circle plus another h.

00:52 So it's going to be 2h plus x plus one half times a circumference of a circle.

01:00 And the circumference of a circle is pi times 2r.

01:04 So it's going to be pi times 2 .r.

01:07 2 r, but r is actually equal to x over 2.

01:11 So let's go ahead and put that in right away.

01:14 So 2 times x over 2.

01:18 And then you'll notice those 2s cancel.

01:20 And we're going to try to keep this as simple as we possibly can.

01:25 So we can simplify this just a little bit more and say 2h plus, let's see, 1 plus pi over.

01:40 Over 2, and that times x.

01:44 And if we want to clean that up even more, we could say, well, one is really 2 over 2.

01:50 And then we'll have a common denominator there.

01:52 So we have that our perimeter 30 feet is equal to 2 times h plus 1, no, 2 plus pi all over 2 times x.

02:08 And the thing we're trying to do is find a formula for the area.

02:13 And then we also want to find the dimensions that'll maximize.

02:16 So the area is going to be the area of the rectangular part.

02:19 We'll call that a1 and the circular part of the semicircular part.

02:25 So it's going to be a1 plus a2.

02:32 And a1, that part's easy.

02:35 That's a rectangle.

02:37 So that's going to equal x times h.

02:40 Say h times x.

02:41 And then a 2, the radius of the circle is x over 2, and pi r squared is the area of a circle, but we only want half of it.

02:52 So we're going to go up one half times pi times r squared.

02:59 We already determined r is actually x over 2.

03:04 So we can clean that up a little bit, and we can say h times x plus pi x squared.

03:13 And 2 squared over here is 4 times 2 is 8.

03:16 So this is now our area.

03:20 But we don't want that h in there.

03:22 We need to replace that with something in terms of x.

03:25 So we're going to come back to the perimeter and solve for h.

03:28 So we're going to say 30 minus 2 plus pi over 2 times x equals 2h.

03:40 And then we'll just divide everything by 2.

03:45 And that will get us.

03:46 What h is.

03:47 So let's clean that up a little bit.

03:49 So 30 divided by 2 is 15.

03:54 And this fraction here divided by 2 is going to be 2 plus pi over a half times 2.

04:03 So 4, it's not a half times 2, a half times a half.

04:08 And then we'll still have our x.

04:12 So that's h.

04:14 And if we're really careful and substitute that in cautiously.

04:19 We should be able to clean this up and get our area function.

04:22 So instead of h, we're going to substitute in our new expression, and we're still going to multiply that by x.

04:30 So we're going to have, let's see, 15 minus 2 plus pi over 4 times x, close of parentheses.

04:45 That's our h and there's our x and then we're still going to have that pi x squared over 8.

04:55 So now let's see what we can do here...

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