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In the following exercises, solve the given maximum and minimum problems.In Example $3,$ change $1600 \mathrm{ft}$ to $2400 \mathrm{ft}$ and then proceed.

Calculus 1 / AB

Chapter 24

Applications of the Derivative

Section 7

Applied Maximum and Minimum Problems

Derivatives

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In mathematics, precalculu…

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In mathematics, a function…

02:38

For the following exercise…

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04:19

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All right. So when this problem, we're trying to minimize the, um, dimension or the amount of material, which is the amount of fencing we need. So we're trying to minimize the perimeter, which is equal to two length plus two with Pierre, right, and were given that the area has to fit a total of 400 feet, so length, times with must be 400. So to roost, reduce the sounds of just one variable. We can set one of these between variables to another using this form form or this equation here. So, for example, we can sit well, we can set link to equal to 400 or sorry, uh, this should be 1600 1600 so this will become 1600 divided by a W so we can plug that here into our, um, into our perimeter. So that becomes 3200 divided by the width, plus two times the with is our perimeter, and we're trying to minimize this. So to minimize we take the derivative said it equal to zero. So the derivative here is going to be negative. 3200 divided by W squared plus two is equal to zero. Okay, so solving groups solving for solving for zero or sorrow, sorrow solving for W we get that 3200 over w squared is equal to two. Cross multiply. You get that w squared is equal to 1600. Or that w is equal to 400. Okay, so we have our with an onion, so this should be, um, in feet, 400 feet here. Okay. Now our length is going to be 1600 divided by 400 or we get that. Our link is also going to be 400 feet. All right, so we have finished our problem.

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