# Optimisation - Critical Numbers for Complex Functions.

Tags:
1. Mar 11, 2015

### Khronos

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.)

3. 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.

2. Mar 11, 2015

### Dick

It's a quadratic equation. There is a formula to find the zeros without factoring. Remember?

3. Mar 11, 2015

### 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: Mar 11, 2015
4. Mar 11, 2015

### Ray Vickson

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.