Maximum Profit: find best selling price per can

[cem348]

I don't know what I'm doing wrong in this problem; it seems pretty straightforward. Thanks so much for any help!

The manufactuer of Brand X Cola has found that his monthly profit P depends on the selling price per can x according to the formula:

. . .P = (10x - 130) times e raised to the -x/30

...where "x" is the selling price in cents per can. If he is unwilling to drop the price below 30 cents a can, what is the maximum feasible profit?

The answer is 43 cents, which I assume refers to the price per can and not the maximum feasible profit.

I've solved a couple different methods and can't get the 43 figure. Integration wouldn't work for this problem, right? I would then need the rate at which its changing. I've tried getting the derivative of P and then setting it equal to zero but I don't get 43. Can anyone help; am I making math errors or solving the problem incorrectly? Thank you!

 

[galactus]

Differentiate your function.

\(\displaystyle \L\\\frac{d}{dx}[(10x-130)e^{\frac{-x}{30}}]=(\frac{43}{3}-\frac{x}{3})e^{\frac{-x}{30}}\)

Now, set that to 0 and solve for x. I think you'll find your answer.

You should be able to see the answer right off. Look inside the parentheses.