# Find math maximum homework

1. May 26, 2004

### Nallen

I've got an assesment task tommorow and one of the questions on the practice exam has got me stumped
The eqaution is y=21xe^(-.9x) - 12xe^(-.6x)

The question asks to show how the maximum can be
e^(0.3x) = 70-63x/40-24x

If differentiated it and taken out the common factors to get
21e^(-.9x)(-.9x + 1) - 12e^(-.6x)(-.6x + 1) = 0

but have no idea how to make it e^(0.3x) = 70-63x/40-24x

If any one could help i would really appreciate it

2. May 26, 2004

### arildno

First of all, write the answer properly:
$$e^{0.3x}=\frac{70-63x}{40-24x}$$
Rewrite the right-hand side as:
$$\frac{70-63x}{40-24x}=\frac{70}{40}\frac{1-0.9x}{1-0.6x}$$

See if you can relate this last expression to the equation giving the critical value for y.