Help applied problem max & min

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In summary, we need to use all of the wire for the square and none for the circle in order to enclose the maximum area.
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Homework Statement



A wire that is 7 feet long is to be used to form a square and a circle. How much of the wire should be used for the square and how much for the circle to enclose the maximum area?

Possible Solutions:
a. o feet for the square and 7 feet for the circle
b. pi/7 feet for the square and 49-pi/7 feet for the circle
c. 7/4+pi feet for the square and 7pi+21/4+pi feet for the circle
d. none of these


Homework Equations



volume?
circunference?


The Attempt at a Solution



differentiate and equal to cero... but I need to find the equation first

please help.. and thanx!
 
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  • #2




Thank you for your question. I am a scientist and I would be happy to help you find a solution to this problem. Firstly, I would like to clarify that the question is asking for the maximum area, not volume. The equations you need to use are those for area and circumference, as you have correctly mentioned.

To find the maximum area, we need to use the formula for the area of a square, A = s^2, and the formula for the area of a circle, A = πr^2. Let's start by representing the wire length, which is 7 feet, as the sum of the four sides of the square and the circumference of the circle. This can be written as:

7 = 4s + 2πr

Next, we need to express the area of the square and the circle in terms of s and r:

As = s^2
Ac = πr^2

Now, we can substitute these expressions into the original equation to get:

7 = 4As^(1/2) + 2πAc^(1/2)

To find the maximum area, we need to differentiate this equation with respect to either s or r, and set it equal to zero. Let's differentiate with respect to s:

dA/ds = 2s^(-1/2) = 0

Solving for s, we get s = 0, which is not a valid solution. Therefore, we need to differentiate with respect to r:

dA/dr = 2πr = 0

Solving for r, we get r = 0, which is also not a valid solution. However, we know that the circumference and the area cannot be equal to zero, so we need to look at the endpoints of the interval. In this case, s and r must be positive, so we can set the derivative equal to zero and solve for s and r:

4s^(1/2) = 0 --> s = 0 (not valid)
2πr^(1/2) = 0 --> r = 0 (not valid)

Therefore, the only valid solution is when s and r are at their maximum values, which means we need to use all of the wire for the square and none for the circle. So, the correct answer is a. 0 feet for the square and 7 feet for the circle.

I hope this helps
 

FAQ: Help applied problem max & min

1. What is an applied problem in science?

An applied problem in science refers to a real-world issue or situation that requires the use of scientific knowledge and methods to find a solution. These problems can cover a wide range of topics, from environmental concerns to medical issues, and often require interdisciplinary approaches to fully understand and address.

2. How is the maximum and minimum used in applied problems?

The maximum and minimum values are commonly used in applied problems to determine the optimal solution. In many cases, these values represent the highest or lowest point of a certain variable, such as cost or efficiency, and can help scientists make informed decisions about the best course of action.

3. What is the role of data analysis in solving applied problems?

Data analysis plays a crucial role in solving applied problems as it helps scientists make sense of large amounts of data and identify patterns and trends. By analyzing data, scientists can gain a better understanding of the problem at hand and make informed decisions about the best solutions.

4. How do scientists determine the maximum and minimum values in applied problems?

There are various methods that scientists use to determine the maximum and minimum values in applied problems. These may include mathematical calculations, experimental studies, and computer simulations. The most appropriate method will depend on the specific problem and available resources.

5. Can applied problems have multiple maximum or minimum values?

Yes, applied problems can have multiple maximum or minimum values. This often occurs when there are multiple variables involved in the problem, and different combinations of these variables can lead to different maximum or minimum values. In such cases, scientists must carefully analyze the data and consider all factors to determine the most optimal solution.

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