# Thread: Real-life application of Calculus

1. ## Real-life application of Calculus

Hello all,

I am currently doing some real-life applications for calculus and stumbled upon the following question, can anyone please help me?

When a chemical factory produces a certain compound they have observed that the daily amount y of defective compounds depends on the total amount x that they produce according to the empirical formula:
y = 0.02x + 0.00005x^2
where x and y are measured in kilograms. The manufacturer earns $100 per kilogram of nondefective compound and incurs a loss of$20 per kilogram of defective compound.
(a) Write down a function for the total profit P(x).
(b) How many kilograms should they produce to maximise their profits?

Any help is much appreciated,

Dranalion

2. I hope I understand your question correctly.
for every xkg of this compoud, ykg is defective. so they lose $20y for every x amount they produce and earn$100(x-y). so their profit is $100(x-y)-$20y. the next find the maximum point of the graph of P(x) by finding the stationary points

3. How do I find the function, P(x)? Does finding the stationary points allow me to do so?

4. Originally Posted by Dranalion
How do I find the function, P(x)? Does finding the stationary points allow me to do so?
P(x) is the profit, that is, amount they earn minus what they lose. so P(x)=100(x-y)-20y

5. How about Part B of the question:

(b) How many kilograms should they produce to maximise their profits?

Any help is much appreciated,

Dranalion

6. Sorry, substituting y = 0.02x + 0.00005x^2 into P(x), the result is:

P(x) = 97.6x + 0.0006x^2

P'(x) = 97.6 + 0.012x (for finding the stationary points, then allowing that to equal zero to find the maximum profits

Is this correct? Am I on the right track?

7. yes, you got it!