Answer to Question #196898 in Calculus for musfirah

Question

The demand function for a firm's product is

q-150,000-75p

where q equals the number of units demanded and p equals the price in dollars. (a) Determine the price which should be charged to maximize total revenue.

(b) What is the maximum value for total revenue? (e) How many units are expected to be demanded?

Answer 1

(a)

$$Revenue = Price \times Quantity.$$

$$R=p(150000-75p)$$

$$R=150000p-75p^2$$

Find the derivative with respect to $p$

$$R'=(150000p-75p^2)'=150000-150p$$

Find the critical number(s)

$$R'=0=>150000-150p=0$$

$$p=1000$$

Critical number $p=1000.$

If $p>1000, p'<0, R$ increases.

If $p<1000, p'>0, R$ decreases.

The function $R$ has a local maximum at $p=1000.$

Since the function $R$ has the only extremum, then the function $R$ has the absolute maximum at $p=1000.$

The price $p=\$1000$ will maximize total revenue.

(b) $R(1000)=150000(1000)-75(1000)^2=\$75,000,000$

(e)

$$q=150000-75(1000)=75,000 (units)$$