Optimization calculus problem

In summary, the total illumination an object receives from two light sources is the sum of the illuminations from each source. To find the distance on the line between the sources where the object receives the least illumination, the derivative of the total illumination expression should be taken and set equal to zero, solving for x.
  • #1
colbenstein
3
0

Homework Statement


The illumination of an object by a light source is directly proportional to the strength of the source and inversely proportional to the square of the distance from the source. If two light sources, one three times as strong as the other, are placed 10 feet apart, where should an object be placed on the line between the sources so as to receive the least illumination?


Homework Equations


1/illumination=distance^2 and illumination=strength. 1/3x=y distance=10 so 1/100= illumination??


The Attempt at a Solution


find the derivative of what equation to optimize?
 
Physics news on Phys.org
  • #2
Let s = strength of a source
Let d = distance from source.

I =ks/d2

The way I read this is that the total illumination an object receives is the sum of the illuminations from the two light sources, so
IT = I1 + I2

From this relationship you should be able to write the total illumination an object at a distance of x feet from the left source gets.

Then differentiate the expression for IT.
 

Related to Optimization calculus problem

1. What is optimization calculus?

Optimization calculus is a branch of mathematics that deals with finding the best possible solution to a problem. This involves using calculus techniques, such as derivatives and integrals, to optimize a given function or system.

2. What is the difference between a local and global maximum/minimum?

A local maximum or minimum refers to the highest or lowest point in a specific region of a function. On the other hand, a global maximum or minimum is the highest or lowest point of the entire function. Local extrema can exist at different points within a function, while global extrema can only exist at one point.

3. How do you find the critical points of a function?

The critical points of a function are the points where the derivative of the function is equal to zero or undefined. To find these points, you can take the derivative of the function and set it equal to zero, then solve for the variable. The solutions to this equation will be the critical points of the function.

4. What is the process for solving an optimization calculus problem?

The process for solving an optimization calculus problem involves the following steps:

  • Identify the objective function, which is what needs to be optimized.
  • Identify any constraints, such as limitations or restrictions on the variables.
  • Use calculus techniques, such as derivatives and integrals, to find the critical points of the objective function.
  • Evaluate the critical points and any endpoints to determine the maximum or minimum value of the function.

5. What are some real-life applications of optimization calculus?

Optimization calculus has many real-life applications, including:

  • Maximizing profits in business by finding the optimal production levels.
  • Minimizing costs in engineering by finding the most efficient design.
  • Maximizing crop yield in agriculture by optimizing fertilizer usage.
  • Minimizing travel time in transportation by finding the shortest route.
  • Maximizing energy efficiency in buildings by optimizing heating and cooling systems.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
2K
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
4K
Replies
6
Views
4K
  • Other Physics Topics
Replies
2
Views
3K
  • Math POTW for University Students
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
839
  • Astronomy and Astrophysics
Replies
1
Views
303
  • Calculus and Beyond Homework Help
Replies
9
Views
3K
Back
Top