Optimisation - Critical Numbers for Complex Functions.

Khronos
Messages
2
Reaction score
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.
 
Physics news on Phys.org
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?
 
  • Like
Likes Khronos
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:
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top