Find math maximum homework

[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 anyone could help i would really appreciate it

 

[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.

 

[M98Ranger]

To find the maximum of a function, you need to find the critical points of the function, where the derivative is equal to 0 or does not exist. In this case, the derivative of y with respect to x is:

y' = 21e^(-0.9x)(-0.9x + 1) - 12e^(-0.6x)(-0.6x + 1)

To find the critical points, set y' equal to 0 and solve for x:

0 = 21e^(-0.9x)(-0.9x + 1) - 12e^(-0.6x)(-0.6x + 1)

Simplifying this equation, we get:

0 = -18.9xe^(-0.9x) + 21e^(-0.9x) - 7.2xe^(-0.6x) + 12e^(-0.6x)

Factoring out e^(-0.6x), we get:

0 = e^(-0.6x)(-18.9x + 21 - 7.2x + 12)

0 = e^(-0.6x)(-26.1x + 33)

Setting each factor equal to 0, we get:

e^(-0.6x) = 0 or -26.1x + 33 = 0

Since e^(-0.6x) can never equal 0, we can ignore that solution.

Solving for x, we get:

x = 33/26.1 = 1.26

Now, to find the maximum, we can plug this value of x back into the original function:

y = 21(1.26)e^(-0.9(1.26)) - 12(1.26)e^(-0.6(1.26))

y = 15.174 - 6.048 = 9.126

Therefore, the maximum value of the function is 9.126, which occurs when x = 1.26.

To show how the maximum can be e^(0.3x) = 70-63x/40-24x, we can substitute this value of x into the original function:

y = 21(1.26)e^(-0.9(1.26))