Answer to Question #92064 in Algebra for Linh

Question

You want to mass produce your boxes from Question 2 to sell. You can make your boxes out of different quality paper (q). The higher the quality the paper the less you will sell but the more you can charge. You have done your market research have come up with the following equations.

n=(100-3q)where n is number of sales,q is quality of paper from 1 to 10.
total sale=$0.5qn
total cost=$0.3qn

Answer 1

$$n=(100-3q)$$

having substituted n in the formulas for sales and cost:

$$sale=0.5q(100-3q)$$

$$cost=0.3q(100-3q)$$

the profit is given by

$$profit=sale-cost=f=0.2(100q-3q^2)$$

the derivative of the profit function with respect to q is

$$t=diff(f)=0.2(100-6q)$$

hence t=0 when q=100/6=50/3>10.

It means one should check f at q=1 and q=10

because on an interval from q=1 to q=10 the function f is monotonous,

q=1 -> total profit=19.4,

q=10 -> total profit=140.

Answer: Maximum total profit=140 when q=10.