Question

the demand for a certain type of cosmetic is given by p = 160 - x, where p is the price in dollars when x units are demanded. (a) find the revenue r(x) that would be obtained at a price p. (hint: revenue = demand × price) (b) graph the revenue function r(x). (c) find the price that will produce maximum revenue. (d) what is the maximum revenue? (a) the revenue function that would be obtained at a price p is r(x) = \square.

Explanation:

Step1: Define revenue function

Revenue = units demanded × price, so $R(x) = x \cdot p = x(160 - x) = 160x - x^2$.

Step2: Identify vertex of quadratic

For $R(x) = -x^2 + 160x$, vertex at $x = -\frac{b}{2a} = -\frac{160}{2(-1)} = 80$.

Step3: Find price for max revenue

Substitute $x=80$ into $p=160-x$: $p=160-80=80$.

Step4: Calculate max revenue

Substitute $x=80$ into $R(x)$: $R(80)=80×80=6400$.

Answer:

(a) $160x - x^2$
(b) A downward-opening parabola with vertex at (80, 6400), x-intercepts at 0 and 160.
(c) $80$
(d) $6400$