Calculus-Optimization Problems

  • Thread starter the_morbidus
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In summary, the study has determined that at a large social party, the number of separate conversations occurring follows the mathematical progression N(t)=30t-t^2, where t is the time in minutes since the party began. The maximum number of interactions occurs at t=15 minutes, with a value of 225 conversations. This was determined by taking the first derivative of the function and setting it equal to 0, then plugging that value back into the original function. The result confirms that at 15 minutes, there are the most conversations occurring at the party.
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the_morbidus
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Homework Statement


A study has determined that interactions at a large social party follow the mathematical progression N(t)=30t-t^2 , where t is the time in minutes since the party began, and N is the number of separate conversations occurring. At what time in a party do the most conversations occur? what is the maximum number of interactions?


Homework Equations


The Product Rule
F(x)=f(x)g(x) then F'(x)=f(x)g'(x)+f'(x)g(x)

The Chain Rule for Polynomials
F(x)=(f(x))^n,then F'(x)=nf'(x)f(x)^n-1


The Attempt at a Solution



So i decided to get the 1st derivative of N(t)=30t-t^2
N'(t)=30-2t
so now I'm trying to find t when the function equals 0.
0=30-2t
-30/-2=t
t=15 so this is the time since the party began.

now this is the part I'm not so sure of myself, i add the value of t to the 1st function
N(15)=30(15)-(15)^2
=450-225
=225 so yeah i don't understand the result, what does this one equal to? I'm just confused on how to obtain the results i need here.
 
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  • #2
225 is N(15). I guess that's the 'maximum number of separate conversations occurring'. I think you've solved the problem.
 
  • #3
lol, see, that is what happens when you doubt yourself all the time like i do hahaha.
 

Related to Calculus-Optimization Problems

1. What is the purpose of optimization problems in calculus?

The purpose of optimization problems in calculus is to find the maximum or minimum value of a function. This is important in many fields such as economics, engineering, and physics where finding the most efficient solution is crucial.

2. How do you set up an optimization problem in calculus?

To set up an optimization problem in calculus, you need to identify the objective function and the constraints. The objective function is the quantity that needs to be maximized or minimized, while the constraints are the limitations or conditions that the solution must satisfy.

3. What are some common techniques used to solve optimization problems in calculus?

Some common techniques used to solve optimization problems in calculus include taking the derivative of the objective function and setting it equal to zero, using the first or second derivative test to determine if it is a maximum or minimum point, and using the method of Lagrange multipliers for constrained optimization problems.

4. Can optimization problems have multiple solutions?

Yes, optimization problems can have multiple solutions. In some cases, there may be more than one maximum or minimum point for the objective function. It is important to check all possible solutions to ensure that the optimal solution is found.

5. How do optimization problems in calculus relate to real-life situations?

Optimization problems in calculus are used to model and solve real-life situations where there is a need to maximize or minimize a certain quantity while taking into account various constraints. For example, finding the most cost-effective way to produce a product or determining the optimal route for a delivery truck.

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