Optimisation - Critical Numbers for Complex Functions.

In summary, the conversation was about finding the temperature at which water has its maximum density using a given formula. The steps taken to solve the problem included rewriting the equation into scientific notation, finding the derivative using the power rule, and looking for critical points. The solution was found to be 3.966514624 degrees, but it is important to note that this may be a minimum instead of a maximum and further testing is needed to confirm. Additionally, the constraints of the problem should also be considered when finding the maximum density.
  • #1
Khronos
2
0
Hi everyone, I need a little bit of help with an optimization problem and finding the critical numbers. The question is a follows:

Question:
Between 0°C and 30°C, the volume V ( in cubic centimeters) of 1 kg of water at a temperature T is given approximately by the formula:
V = 999.87 − 0.06426T + 0.0085043T2 − 0.0000679T3

Find the temperature at which water has its maximum density. (Round your answer to four decimal places.)

The Attempt at a Solution



I have done the following steps:
Re-write equation into scientific notation:
V(t)=-6.79x10-5T3+8.5043x10-3T2-6.426x10-2T+999.87

Found Derivative using Power Rule:
V'(t)=-2.037x10-4T2+1.70086x10-2T-6.426x10-2

I need to find critical points, where V'(t)=0 or V'(t)=und.
There are no obvious factors and I need help to find the zero's of the derivative.
Any help would be greatly appreciated.
 
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  • #2
Khronos said:
Hi everyone, I need a little bit of help with an optimization problem and finding the critical numbers. The question is a follows:

Question:
Between 0°C and 30°C, the volume V ( in cubic centimeters) of 1 kg of water at a temperature T is given approximately by the formula:
V = 999.87 − 0.06426T + 0.0085043T2 − 0.0000679T3

Find the temperature at which water has its maximum density. (Round your answer to four decimal places.)

The Attempt at a Solution



I have done the following steps:
Re-write equation into scientific notation:
V(t)=-6.79x10-5T3+8.5043x10-3T2-6.426x10-2T+999.87

Found Derivative using Power Rule:
V'(t)=-2.037x10-4T2+1.70086x10-2T-6.426x10-2

I need to find critical points, where V'(t)=0 or V'(t)=und.
There are no obvious factors and I need help to find the zero's of the derivative.
Any help would be greatly appreciated.

It's a quadratic equation. There is a formula to find the zeros without factoring. Remember?
 
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Likes Khronos
  • #3
Oh dear... of course!... I don't know why I didn't think of that. Thank you so much haha.

The minimum of the function was 3.966514624 degrees for anyone interested.
 
Last edited:
  • #4
Khronos said:
Oh dear... of course!... I don't know why I didn't think of that. Thank you so much haha.

The minimum of the function was 3.966514624 degrees for anyone interested.

You were told to find the maximum, while you claim to have found the minimum.

Have you tested whether your point is a maximum or a minimum? Just setting ##V'(T) = 0## will not tell you this; you need to use a second-order test (involving the second derivative ##V''(T)##), or use some other types of tests.

Also: how do you know you should set the derivative to 0 at all? Perhaps the constraints ##0 \leq T \leq 30## mess things up? You need to check that as well.
 

Related to Optimisation - Critical Numbers for Complex Functions.

1. What are critical numbers in optimization and why are they important?

Critical numbers are points on a function where the derivative is equal to zero or does not exist. They are important in optimization because they represent the potential maximum or minimum values of the function. By finding these critical numbers, we can determine the optimal solution for a given problem.

2. How do you find critical numbers for complex functions?

To find critical numbers for complex functions, we first take the derivative of the function using the rules of differentiation. Then, we set the derivative equal to zero and solve for the variable(s). The resulting values are the critical numbers for the function.

3. What is the difference between local and global critical numbers?

Local critical numbers refer to the critical numbers that are only valid within a specific interval of the function. These points may represent local maximum or minimum values. On the other hand, global critical numbers are valid for the entire function and represent the overall maximum or minimum values.

4. Can critical numbers be negative or imaginary?

Yes, critical numbers can be negative or imaginary. In fact, complex functions often have critical numbers that are imaginary. These numbers are still important in optimization as they can represent either a local or global maximum or minimum value of the function.

5. How are critical numbers used in real-world applications?

Critical numbers are used in real-world applications to optimize various processes, such as finding the optimal production level in a manufacturing plant or determining the most efficient route for a delivery truck. By analyzing the critical numbers of a complex function, we can make more informed decisions and improve the overall efficiency of a system.

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