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Consider the following problem: A farmer with $ 750ft $ of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens?(a) Draw several diagrams illustrating the situation, some with shallow, wide pens and some with deep, narrow pens. Find the total areas of these configurations. Does it appear that there is a maximum area? If so, estimate it.(b) Draw a diagram illustrating the general situation. Introduce notation and label the diagram with your symbols.(c) Write and expression for the total area.(d) Use the given information to write an equation that relates the variables.(e) Use part (d) to write the total area as a function of one variable.(f) Finish solving the problem and compare the answer with your estimate in part (a).

a) max area appears to be $14,062.5 \mathrm{ft}^{2}$b) Let $x$ be one side of the overall pen and $y$ the other\c)c) $A=y \times x$d) $5 x+2 y=750$e) $A=x\left(375-\frac{5}{2} x\right)$f) Maximum area is 14062.5 $\mathrm{ft}^{2}$

05:59

Wen Z.

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 7

Optimization Problems

Derivatives

Differentiation

Volume

Baylor University

University of Michigan - Ann Arbor

University of Nottingham

Boston College

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So we're gonna start off with a formula where we can pick a value for width and get a value for the length of the other sides. So it's going to be 750 minus two W um, And since we have 750 ft of fencing to work with, we subtract two w from it to get what's left for the other sides. Um and then, since there are five parallel sides, what we have is that P is equal to 7. 50 minus two w over five. So we could alternatively, make a formula where we pick the length of the five parallel sides instead. But this is what our fencing is gonna look like. So, um, we have 200 ft here, 70 ft here, and then it will be broken up like this. Um, that would give us 70 times 200 which equals 14,000 ft squared. There's other drawings. You could do it there. Give us 9000, but experimenting with numbers. The max area we seem to get is somewhere around here. But we want to figure this out for ourselves. We want the absolute maximum. So for 11 B, part B. We're gonna let one side of the overall pen B X and the other side B y. So it's kind of like Lake. Yes. So this is going to be hard. Why? This will be X. And then there will be the rest of the pen and all of it subdivisions. Then we let it be the X width and the y length and a area so a is going to equal X times y. So then, for part that was C. So for part D, Um, what we have is letting x be the length of the end wall and y be the length of the dividing wall. What we end up having is two x plus three, uh, two x plus three x So this is the two X. And then there's the three acts right here. 123 and then we have two y right here and right here. So plus two y is going to echo five x plus two y and we know that five X plus two y is going to add up to equal +17 50 because that is our amount of fencing that we have. So now we can solve for y and we get y equals 750 minus five x over two. And we know that a equals x times y so we can substitute that in. So we get a equals X times this portion so we can just rewrite that as 375 x minus five house x squared. Then on top of that, um, for F, what we can do is differentiate this area and we end up getting a prime being equal to 375 minus five x. We set this equal to zero and see, um, that X equals 75. So what that tells us is that since when a prime of X is less than zero, you get when X is greater than 75 and a prime of X is greater than zero when X is less than 75. What we see is that X equals 75 is a local max. In that area, a of 75 is going to equal 14,062 750.5 ft squared

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