Is there a way to prove that A(x) is minimum when x=2r?

In summary, the conversation discusses the relationship between the perimeter and area of a circle and square, where the sum of their perimeters is a constant k. It is proven that the sum of their areas is minimized when the side of the square is double the radius of the circle. This is achieved by finding the critical point and using the second derivative to show that it is a minimum point. By setting up an expression for the sum of the areas, A, and eliminating one variable using the given condition, A can be expressed as a function of one variable, such as A(x).
  • #1
Rush123
1
0
The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the
sum of their areas is least when the side of square is double the radius of the circle.
is this way correct?
assume that x is side of square and r is radius of circle
k=4x+2∏r

sum of areas=area of square+area of circle
find critical pt and find 2nd derivative to show its min.
 
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  • #2
Yes, that will work. What do you get when you that?
 
  • #3
Let
A(x,r) be the sum total of the areas. Can you set up the expression for A?

Furthermore, by utilizing the condition that the sum of the perimeters equals k, can you eliminate one of the variables in A, so that A depends on one of the variables only, for example as A(x)?
 

Related to Is there a way to prove that A(x) is minimum when x=2r?

What is a maxima and minima problem?

A maxima and minima problem is a type of mathematical optimization problem where the goal is to find the maximum or minimum value of a function.

What is the difference between a local maxima and global maxima?

A local maxima is the highest point in a specific interval of a function, while a global maxima is the highest point in the entire domain of a function.

How do you find the critical points of a function?

To find the critical points of a function, you need to take the derivative of the function and set it equal to zero. The values that satisfy this equation are the critical points.

What is the first derivative test?

The first derivative test is a method used to determine whether a critical point is a maximum or minimum. If the derivative of the function is positive to the left of the critical point and negative to the right, it is a local maxima. If the derivative is negative to the left and positive to the right, it is a local minima.

What is the second derivative test?

The second derivative test is a method used to determine whether a critical point is a maximum or minimum. If the second derivative of the function is positive at the critical point, it is a local minima, and if it is negative, it is a local maxima.

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