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Show that the answer to Exercise 3 follows from Exercise 2.

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 4

Applications I - Geometric Optimization Problems

Derivatives

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Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

01:43

Confirm your answer for Ex…

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Use a graphing calculator …

01:11

In Exercises $3-8,$ explai…

00:34

Express the patterns of th…

these four problems together since they're all pretty similar And in fact they refer to one another a bit. Um so the first step is to find the dimensions of a largest rectangle whose perimeter is 144 ft. So the perimeter of a rectangle that has side A. And B. Is to A plus to be. Um And then the area is of course A times B. So we can solve this for A. And we get A. Is 70 to minus B. Plug that into here. So now we have a as a function just to be, you can take the derivative to uh find the maximum. So we get to deliver their with 72 minus um to be selling that equals zero. So for B. One B one equals 36. And if we come back up here plug that in there, You see that a one is also 36. So we get a square now um problem to basically just says that um we always should get a square no matter what the perimeter is. So we just generalized this problem up here. So that now 144, I just called P for the Perimeter going through the same steps, we have a equals P over two minus B times B. So for a year plugged into there, take the derivative, we wind up with P over two minus to be setting that equal to zero. We get the critical point and that is the one equals P over four and we plug that back into here and we get a one equals P over four. So it's always it's always equal to one another And always 1/4 of their perimeter. Now let's see they ask us um the sum of two positive numbers is 100 and should be the number. And what should the numbers be if the product is the the largest as possible? All right. So we have a plus B equals 10 and then the some of the product is £8 B. And we have so solving this for a plugging it in here, we have y equals 10 10 minus B. Times C squared. Taking the driver that we get 10 minus to be set this equal to zero. So it would be um substitute and be one set the secret zero. And we get the one he was five. And plugging that back in here. A 1.5. And this is sure that this answer follows from, you know, from this. Well, we can see here the form is exactly the same. Right? So if we said if we set pe for the 20 here, we would actually wind up having this exact same problem. And so, you know, we would get five case. So it's really basically just the exact same. Some problem is just, you know, one is uh kind of a depends on geometry. And this is just, you know, just general generic math problem here

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