Max-Min problem - solve only with calculus

  • Thread starter Ara macao
  • Start date
  • Tags
    Calculus
In summary, the maximum possible area for a triangle inscribed in a circle of radius r is achieved when the triangle is isosceles. This can be proven by showing that the third vertex of the triangle must lie on the perpendicular bisector of the chord, resulting in two congruent right triangles and thus an isosceles triangle.
  • #1
Ara macao
27
0
What is the maximum possible area for a triangle inscribed in a circle of radius r?

So the triangle must be an isosceles one. How do we know that, and how do we prove that? And then how should we proceed?

Thanks!

the scarlet macaw
 
Physics news on Phys.org
  • #2
Two points on the circle define a chord. A third point defines a triangle. Use the chord as the base and the third point will give the maximum area when it's perpedicular distance from the chord (height of triangle) is a maximum. You should have no trouble determining that the maximum height occurs when the third point is on the perpendicular bisector of the chord - i.e. it is an isosceles triangle.

The rest is trivial using the vertex angle at the third point as your variable.
 
  • #3
how do you prove the isosceles-ness of the triangle? I am thinking about trying to set dh/d(theta) = 0 somehow, but other variables keep on getting into the way. Sorry, it may seem clear to y'all, but I am very weak in proofs.
 
  • #4
As I indicated before, you will find the third vertex of the triangle to lie on the perpendicular bisector of the chord. Draw a line from the vertex to the base (chord). That line bisects the chord so you can see two right triangles. Their bases are identical and the line you just drew is common to both. Therefore, the two triangles are congruent so your original triangle is isosceles.
 

What is the Max-Min problem?

The Max-Min problem is a type of optimization problem in calculus where the goal is to find the maximum or minimum value of a function within a given interval. This can be applied to various real-world scenarios, such as finding the optimal production level for a business or the highest point on a roller coaster track.

How do you solve the Max-Min problem using calculus?

To solve the Max-Min problem using calculus, you first need to find the critical points of the function by taking its derivative and setting it equal to zero. Then, you can use the first or second derivative test to determine whether the critical point is a maximum or minimum. Finally, you can check the endpoints of the interval to see if they provide a higher or lower value than the critical point, and that will give you the maximum or minimum value of the function.

What is the first derivative test in the Max-Min problem?

The first derivative test is a method used to determine whether a critical point of a function is a maximum or minimum. It involves evaluating the sign of the derivative at the critical point. If the derivative is positive, the critical point is a minimum, and if the derivative is negative, the critical point is a maximum.

What is the second derivative test in the Max-Min problem?

The second derivative test is another method used to determine whether a critical point of a function is a maximum or minimum. It involves evaluating the sign of the second derivative at the critical point. If the second derivative is positive, the critical point is a minimum, and if the second derivative is negative, the critical point is a maximum. If the second derivative is zero, then the test is inconclusive, and another method must be used.

What are some real-world applications of the Max-Min problem?

The Max-Min problem has various real-world applications, including finding the optimal production level for a business, determining the maximum or minimum cost for a project, and finding the highest or lowest point on a curve. It can also be applied to economics, engineering, and physics, among other fields.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
774
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
923
  • Computing and Technology
Replies
13
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
792
Replies
8
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
967
  • Calculus and Beyond Homework Help
Replies
24
Views
3K
  • Calculus and Beyond Homework Help
Replies
1
Views
962
  • Calculus
Replies
3
Views
1K
Back
Top