Maximum Profit Application Problem

[cmpolasek]

A company manufactures digital cameras. The company estimates that the profit from camera sales is
P(x) = -0.02x^(3)+0.01x^(2)+1.2x-1.1
Where P is the profit in millions of dollars and x is the amount in hundred-thousands of dollars spent on advertising. Determine the maximum profit that can be achieved and the amount of advertising that needs to be spent to achieve that maximum profit. Round both answers to the nearest thousand.

Now, I am not sure where to even begin, Ive done maximum volume and area applicatin problems before but nothing like this, and the text book does not provide any problems similar to this.

If I could have the step by step process on what to do that woud be incredibly helpful, or you can change the numbers up so its not the same answer but still the same problem.

 

[galactus]

This is just a matter of differentiating the given function.

Differentiate, set to 0 and solve for x. The x will be the amount of advertising for max profit. Plug this value back into the original function will give the max profit.

 

[cmpolasek]

I still do not understand how to solve this equation if you could possibly go into further detail i would appreciate it.

 

[galactus]

If you have done other optimization problems, this should be easy.

I am going to change those decimals to fractions.

Differentiate \(\displaystyle P(x)=\frac{-1}{50}x^{3}+\frac{1}{100}x^{2}+\frac{6}{5}x-\frac{11}{10}\)

Rather straightforward derivative.

Set the derivative to 0 and solve for x

This resulting x value will be the advertising expenditures that give the max profit.

To find this max profit, plug that value back into P(x).

Graph P(x). That will help see the high point of the function.

When you take the derivative of P(x), you get a concave down parabola. Where this parabola cross the x-axis is the values that give your max.

This is because this parabola represents the derivative of P(x). Setting this to 0 and solving for x is where it crosses the x-axis.