# Homework Help: Finding the minimum

1. May 22, 2010

### Procrastinate

Calculate the minimum of PT = 16.8e0.0697t + 204e-0.356t

PT' = 0.0697 x 16.8e0.0697t + -.356 x 204e-0.356t

Stationary points occur when PT'=0. Therefore:

0=0.0697 x 16.8e0.0697t + -.356 x 204e-0.356t

My final answer was that t = 4.29. However, the Graphics Calculator begs to differ and says that it is 5.695.

Attached is my working; where have I gone wrong?

#### Attached Files:

• ###### scan0013.jpg
File size:
17.4 KB
Views:
118
Last edited: May 22, 2010
2. May 22, 2010

### cronxeh

Last edited by a moderator: Apr 25, 2017
3. May 22, 2010

### Procrastinate

Was that using the natural logarithm or an exponent?

Last edited by a moderator: Apr 25, 2017
4. May 22, 2010

### cronxeh

Up to this point you were doing well, then you decided to start dividing and made an error. Try by multiplying it out first then dividing:

0.356*204*exp(-0.356*t) = 0.0697*16.8*exp(0.0697*t)

It should be

72.624*exp(-0.356*t) = 1.17096*exp(0.0697*t)
62.020*exp(-0.356*t)=exp(0.0697*t)

62.020=exp(0.0697*t)/exp(-0.356*t)

ln(62.020)=ln(exp(0.4257*t))

4.1275 = 0.4257*t

t=9.6957

Last edited: May 22, 2010
5. May 22, 2010

### hgfalling

In your work, you seem to have changed 0.30 .... into 0.030.... a little below the middle of the page.

6. May 22, 2010

### Procrastinate

I did that and now I get 9.695...

7. May 22, 2010

### Procrastinate

I rectified that afterwards. Perhaps what I graphed into my Graphics Calculator is wrong.

8. May 22, 2010

### Procrastinate

It is. Silly me.