{Calculus 1} Optimization Problem

In summary, the problem involves dividing a wire into two parts, one shaped into a square and the other into a circle. The objective is to minimize the total area of both shapes by finding the ratio of the circumference of the circle to the perimeter of the square. The area of the circle was incorrectly calculated, leading to a wrong solution. After fixing the equation, taking the derivative and setting it to zero, the correct solution of pi/4 was obtained.
  • #1
youngstudent16
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1

Homework Statement


A wire is divided into two parts. One part is shaped into a square, and the other part is shaped into a circle. Let r be the ratio of the circumference of the circle to the perimeter of the square when the sum of the areas of the square and circle is minimized. Find r.

Homework Equations


Derivative and area of circle and square

The Attempt at a Solution



So I figure I have a wire y since I don't know the length cut it into two pieces x for the square and y-x for the circle. Area of the square is (x/4)^2 While y-x=2pir, r=y-x/2pi So area of circle is (y-x/2pi)^2

Now I want to minimize the total area by taking the derivative and setting it equal to zero. My problem is the y and this doesn't seem right. Any help is appreciated I still struggle with these word problems
 
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  • #2
What is the objective function?

What is the constraint function?

Define those first.
 
  • #3
Dr. Courtney said:
What is the objective function?

What is the constraint function?

Define those first.

I messed up the area of the circle

so I want to minimize
area of square + area of circle = x^2/16 + (y-x)^2/pi

Which would be my constraint

Objective function

In order to get x and y for r = circumference of circle / perimeter of square = (y - x) / x

Do I have all the information I need now my problem is if I take the derivative of that and set it equal to zero I have two variables
 
Last edited:
  • #4
youngstudent16 said:
I messed up the area of the circle

so I want to minimize
area of square + area of circle = x^2/16 + (y-x)^2/pi

Which would be my constraint

Objective function

In order to get x and y for r = circumference of circle / perimeter of square = (y - x) / x

Do I have all the information I need now my problem is if I take the derivative of that and set it equal to zero I have two variables

(1) You do not have two variables; a poor choice of notation has allowed you to mislead yourself. The wire is given, so its length is not variable. Of course, you do not happen to be told its exact length, but for purposes of optimization that is irrelevant: the length is an input parameter, not a variable. Or, to put in another way: if the length of the wire is also regarded as a variable, the best solution is to take length = 0, as that certainly minimizes the total of both areas.
(2) In optimization, the objective function is the thing you are trying to maximize or minimize, while constraints are restrictions on the variables.
(3) Your formula for the area of the circle is wrong.
 
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Likes youngstudent16
  • #5
Yes thank you your first point was my biggest mistake I was making. Its funny how understanding the question correctly makes the process much easier.

I fixed the area equation took the derivative set it equal to zero and finally got the solution of pi/4 as the correct answer
 
  • #6
youngstudent16 said:
Yes thank you your first point was my biggest mistake I was making. Its funny how understanding the question correctly makes the process much easier.

I fixed the area equation took the derivative set it equal to zero and finally got the solution of pi/4 as the correct answer

Correct!
 

Related to {Calculus 1} Optimization Problem

1. What is an optimization problem in Calculus 1?

An optimization problem in Calculus 1 involves finding the maximum or minimum value of a function given certain constraints. This can be solved using techniques such as finding critical points or using the first or second derivative tests.

2. How do you approach solving an optimization problem?

To solve an optimization problem, you must first identify the objective function and any constraints. Then, you can use techniques such as setting up an equation and finding critical points, or using the first or second derivative tests to find the maximum or minimum value.

3. What is the difference between a local and global optimum?

A local optimum is the highest or lowest point within a small interval, while a global optimum is the highest or lowest point of the entire function. In optimization problems, it is important to determine whether the solution is a local or global optimum.

4. Can an optimization problem have multiple solutions?

Yes, an optimization problem can have multiple solutions. This can happen when there are multiple critical points or when the objective function has more than one global optimum.

5. How do you know if a solution to an optimization problem is valid?

A solution to an optimization problem is valid if it satisfies all the given constraints and if it is the maximum or minimum value of the objective function. It is important to check that the solution meets all the requirements of the problem to ensure its validity.

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