Help applied problem max & min

[ayrestaurant]

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!

 

[mighty2000]

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