Minimizing Area between two functions, one which is constant.

In summary, the conversation discusses how to graph the curves y = (x+4)(x-5) and y = ax+2 for various values of a and determine the value of a that minimizes the area between the two curves. It is suggested to find the values of x where the two curves intersect, do the integral between those limits, and take the derivative with respect to a and set it equal to zero. Some suggest using Maple or other software to solve the problem. The final answer is a = -1.
  • #1
solomar
6
0

Homework Statement



Graph the curves y = (x+4)(x-5) and y = ax+2 for various values of a. For what value a is the area between the two curves a minimum?

Homework Equations



y = (x+4)(x-5)
y = ax+2

The Attempt at a Solution



For my attempt I attempted to find the primary equation to optimize. I used Area =
[tex]\int(ax+2-(x+4)(x-5))[/tex]
Since when you graph the two ax+2 is the upper function.
I cannot find a secondary equation to bring it to one variable.

The answer itself is a = -1 (guess and check haha) I am just unsure of how to properly go about minimizing this...
a is what is changing in order for the area to be minimized I understand, and I've exhausted all experimentation with finding a secondary equation in my mind.

Any help or input would be greatly appreciated! This problem is already done on my homework (mymathlab so only the answer is needed), I just really would like to figure out how to properly do this!
 
Physics news on Phys.org
  • #2
To solve that problem you have to find the values of x where the two curves intersect in terms of a (use the quadratic equation). Then do the integral between those limits. Finally take the derivative with respect to a and set it equal to zero. It's really quite a mess.
 
  • #3
Dick said:
To solve that problem you have to find the values of x where the two curves intersect in terms of a (use the quadratic equation). Then do the integral between those limits. Finally take the derivative with respect to a and set it equal to zero. It's really quite a mess.

That's for sure! It's a perfect problem for Maple or its ilk. Maple verifies the OP's intuition that a = -1. Maybe it was a Maple type llab exercise in the first place.
 

Related to Minimizing Area between two functions, one which is constant.

1. What is the purpose of minimizing the area between two functions?

The purpose of minimizing the area between two functions is to find the smallest possible region enclosed by the two functions. This can be useful in various applications, such as optimization problems, to find the most efficient solution.

2. How do you determine which function is constant when minimizing the area between two functions?

The constant function can be identified by observing that it has the same value at all points along the x-axis. This means that its graph is a horizontal line on the coordinate plane.

3. What mathematical techniques can be used to minimize the area between two functions?

Some common mathematical techniques used to minimize the area between two functions include integration, differentiation, and setting up equations to find the critical points or points of intersection between the two functions.

4. Can the area between two functions ever be negative?

No, the area between two functions cannot be negative. The area is always a positive value, representing the region enclosed by the two functions on the coordinate plane.

5. How does minimizing the area between two functions relate to finding the shortest distance between two points?

Minimizing the area between two functions can be thought of as finding the shortest distance between two points, as the shortest distance between two points is the line that connects them, which can be represented as the area under the curve of two functions. By minimizing this area, we are finding the shortest distance between the two points.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
475
  • Calculus and Beyond Homework Help
Replies
1
Views
571
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
414
  • Calculus and Beyond Homework Help
Replies
27
Views
773
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
977
  • Calculus and Beyond Homework Help
Replies
8
Views
499
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
Back
Top