# Real-life application of Calculus

• Oct 1st 2009, 06:55 PM
Dranalion
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
• Oct 1st 2009, 07:25 PM
arze
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
• Oct 2nd 2009, 04:50 AM
Dranalion
How do I find the function, P(x)? Does finding the stationary points allow me to do so?
• Oct 2nd 2009, 11:44 PM
arze
Quote:

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
• Oct 3rd 2009, 10:00 PM
Dranalion
How about Part B of the question:

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

Any help is much appreciated,

Dranalion
• Oct 3rd 2009, 11:24 PM
Dranalion
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?
• Oct 4th 2009, 12:03 AM
arze
yes, you got it!