Question

A manufacturer can produce printer paper at a cost of $2 per ream. The paper has been

selling for $5 per ream, and at that price, consumers have been buying 4,000 reams a month.
The manufacturer is planning to raise the price of the paper and estimates that for each $1
increase in the price, 400 fewer reams will be sold each month. Express the manufacturer’s
monthly profit as a function of the price at which the reams are sold.

Answer 1

Let x be the price per ream in dollars. Then the manufacturer's monthly revenue is given by x times the number of reams sold, which is given by 4000 - 400(x - 5).
The monthly cost is $2 per ream times the number of reams produced, which is given by 4000.
Therefore, the manufacturer's monthly profit as a function of the price x is given by:

Profit = x * (4000 - 400(x - 5)) - 2 * 4000 = 4000x - 400x^2 + 8000 - 8000 = -400x^2 + 4800x - 8000

So the profit is a quadratic function of the price x, with a maximum value at the price that maximizes the profit

Answer 2

Let x be the price per ream in dollars. Then, the manufacturer's profit per ream sold is (x - $2).

The manufacturer estimates that for each $1 increase in the price, 400 fewer reams will be sold each month. Thus, the number of reams sold can be expressed as 4000 - 400(x - $5) reams per month.

The manufacturer's total monthly profit is given by the product of the number of reams sold and the profit per ream, so it can be expressed as:

P(x) = (4000 - 400(x - $5)) * (x - $2)

So the monthly profit as a function of the price per ream is:

P(x) = 4000x - 8000 - 400x^2 + 2000x + 8000

This is a quadratic equation and its maximum value can be found using techniques from algebra or calculus.

Explanation:

ABOVE