Question

In a class of 120 students, everybody would take two hamburgers if the price were zero, and no one would buy hamburgers if the price were $4 or more. assume that the class's demand curve for hamburgers is linear, and give a formula (a) (3 pts) describing quantity as a function of price (b) (3 pts) describing price as a function of quantity (c) (3 pts) total revenue is a function of quantity and can be computed by r = price ×quantity. find the total revenue function corresponding to the demand curve, and conclude that it is quadratic. (d) (3 pts) at what value does its maximum or minimum occur? (use the technique of completing the square.)

Answer 1

Let p be the price and q be the quantity of hamburgers.

a)When price is 0, every one takes 2 hamburgers so, q=120*2 = 240

When p=4, q =0

Hence slope of demand line =$\frac{240-0}{0-4} = \frac{236}{-4} = -60$

Then the demand equation is q=-60p+b ( assume p on x-axis and q on y-axis like y=mx+b)

To find b, we plugin the point (240,0)

    240 = m*0+b

  b= 240

Hence demand equation is q=240-60p.

b) To find p in terms of q,we will solve for p from demand equation.

q-240 = -60p

$\frac{q-240}{-60} = p$

$p=\frac{q}{-60} -\frac{240}{-60}$

$p=4-\frac{q}{60}$

c) Revenue = price * quantity

                   = $p*(240-60p) = 240p-60p^{2}$

Hence it is quadratic equation.

d) $R=-60*(p^{2} -4p)$

             = $-60(p^{2}-4p+4-4)$

             = $-60((p-2)^{2} -4)$

             = $-60(p-2)^{2}+240$

The above equation is a inverted u-shape parabola.

Hence maximum revenue occurs at vertex that is at p=2.