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A 12 inch piece of wire is to be cut into two pieces. One piece is to be used to form a square and the other to form a circle. How should the wire be cut if the sum of the areas is to be maximized?

$\frac{48}{\pi+4}$ in. for square, $\frac{12 \pi}{\pi+4}$ in for circle

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 4

Applications I - Geometric Optimization Problems

Derivatives

Missouri State University

Campbell University

Harvey Mudd College

Boston College

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.

03:29

A piece of wire 12m long i…

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piece of wire long is cut …

08:52

A 24 -in. piece of wire is…

01:21

A person would like to cut…

08:20

A wire of length $12 \math…

08:49

Wire of length 12 $\mathrm…

03:46

A piece of wire $100 \math…

07:21

Circle and square A piece …

07:50

Yeah, We have a total, we have 12" or one ft of wire and we're going to cut it into two pieces, one of length a and one of length b. And we're gonna make a circle out of the one of length a and a square out of the one of length b. And we want to maximize the total area contained in these two objects. And so the area in the circle, well we know the circumference is a, so the radius is a over two pi. So the area is pi r squared or a squared over four pi. And then we know that this is each side has a length of the over four here, So the total length is four is beef. And so the area is B squared over 16. So the total area we get is this um solve this for a A and substituting. And so we get the area is a function of B. Is given by this expression, you can take the derivative of that, that be equal to be one after some simplification, we get four plus pi has be one minus 48. All over eight pi square or a pie. And so what we get is to set that equal to zero. We went up with the one in course 48/4 pi. And that winds up being 6.7 inches. You plug that back into here, we get a the optimal length for a then is 12 pi over four plus pi and that's about uh 5.3 inches. And if we plug that's all back into the area here, that that area is 36 all over four pi or about five square inches. So we're cutting a little more, that we're going to make the make a little more, uh use the longer section to make square and the shorter steps in to make the circle.

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