# Optimization problem (minimization)

• togame
In summary, the problem involves finding the radius of a sector of a circle with a given area, where the perimeter of the sector is the smallest possible. After setting up the equations for area and perimeter, it is determined that the perimeter must be the length of the arc plus the length of the two radii. By setting this up as an equation and finding the derivative, the radius can be solved for and the minimum perimeter can be found.
togame

## Homework Statement

I have a sector of a circle with area 12 square meters. If radius r and angle $\theta$ are chosen so that the that perimeter of the sector is the smallest possible, then what is the radius?

## Homework Equations

I have area of sector as $A=\frac{\theta r^2}{2}$ which is 12
and the length of the arc as $L=\theta r$

## The Attempt at a Solution

my attempt is as follows:
since i am attempting to minimize the circumference, i will need to minimize that function.
$$L=\theta r$$
and since i only want one variable, i use the area function and solve in terms of r and replace it in the circumference function.
$$12=\frac{\theta r^2}{2}$$
$$\theta=\frac{24}{r^2}$$
$$C=r\frac{24}{r^2}$$
$$C=\frac{24}{r}$$
now i need to take the derivative of this function and solve set equal to 0 to get r
$$C\prime=-\frac{24}{r^2}$$
now, this is where i am stuck. since r is squared, this derivative can never be 0, so i believe i am missing a step somewhere or am just confused about how to set it up. any help would be greatly appreciated.

The only thing I can think of is that what they mean by perimeter is the length of the arc + the length of the two radii.

Think about it for a moment, I believe your answer for the derivative makes sense. If the area of the sector remains constant, the larger your radius gets, while keeping the area constant, the smaller the length of the arc gets. Take r to infinity and you get, basically, 0 for the length of the arc. That's why I believe they mean the length AROUND the sector, not just the arc.

Try setting $L = 2r + \theta r$ and see what answer you get!

scurty said:
The only thing I can think of is that what they mean by perimeter is the length of the arc + the length of the two radii.

Oh man, I can't believe I didn't see that! That got me to where I needed to be. Thanks for your help!

## 1. What is an optimization problem?

An optimization problem is a mathematical problem that involves finding the best solution among a set of possible solutions. It typically involves maximizing or minimizing a certain quantity, such as cost, time, or energy, while satisfying certain constraints or limitations.

## 2. What is the objective function in an optimization problem?

The objective function is the mathematical expression that represents the quantity that needs to be minimized or maximized in an optimization problem. It is typically denoted by f(x) or J(x), where x is the variable being optimized.

## 3. How do you determine the constraints in an optimization problem?

Constraints in an optimization problem are the limitations or conditions that the solution must satisfy. They can be determined by considering any physical, practical, or mathematical restrictions that apply to the problem. These can include things like resource limitations, physical boundaries, or mathematical inequalities.

## 4. What is the difference between a local and global minimum in an optimization problem?

A local minimum is a solution that is the lowest point in a specific region of the solution space, while a global minimum is the lowest point among all possible solutions. In other words, a local minimum is the best solution within a limited area, while a global minimum is the overall best solution for the entire problem.

## 5. What are some common methods used to solve optimization problems?

Some common methods for solving optimization problems include linear programming, gradient descent, genetic algorithms, and simulated annealing. These methods use different approaches and techniques to find the best solution for a given problem, and the most appropriate method will depend on the specific problem at hand.

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