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](https://tex.z-dn.net/?f=%5Cfrac%7B240-0%7D%7B0-4%7D%20%3D%20%5Cfrac%7B236%7D%7B-4%7D%20%20%3D%20-60)
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](https://tex.z-dn.net/?f=%5Cfrac%7Bq-240%7D%7B-60%7D%20%3D%20p)
![p=\frac{q}{-60} -\frac{240}{-60}](https://tex.z-dn.net/?f=p%3D%5Cfrac%7Bq%7D%7B-60%7D%20-%5Cfrac%7B240%7D%7B-60%7D)
![p=4-\frac{q}{60}](https://tex.z-dn.net/?f=p%3D4-%5Cfrac%7Bq%7D%7B60%7D)
c) Revenue = price * quantity
= ![p*(240-60p) = 240p-60p^{2}](https://tex.z-dn.net/?f=p%2A%28240-60p%29%20%3D%20240p-60p%5E%7B2%7D)
Hence it is quadratic equation.
d) ![R=-60*(p^{2} -4p)](https://tex.z-dn.net/?f=R%3D-60%2A%28p%5E%7B2%7D%20-4p%29)
= ![-60(p^{2}-4p+4-4)](https://tex.z-dn.net/?f=-60%28p%5E%7B2%7D-4p%2B4-4%29)
= ![-60((p-2)^{2} -4)](https://tex.z-dn.net/?f=-60%28%28p-2%29%5E%7B2%7D%20-4%29)
= ![-60(p-2)^{2}+240](https://tex.z-dn.net/?f=-60%28p-2%29%5E%7B2%7D%2B240)
The above equation is a inverted u-shape parabola.
Hence maximum revenue occurs at vertex that is at p=2.