Answer:
(a)[tex]D(q)=\frac{-1}{25} q+148[/tex]
(b)[tex]S(q)=\frac{1}{50}q+58[/tex]
(c)[tex]p_{*} =88\\\\q_{*} =1500[/tex]
Step-by-step explanation:
(a) For the demand equation D(q) we have
P1: 138 Q1: 250
P2: 108 Q2: 1000
We can find m, which is the slope of the demand equation,
[tex]m=\frac{p_{2} -p_{1} }{q_{2} -q_{1} }=\frac{108-38}{1000-250} =\frac{-30}{750}=\frac{-1}{25}[/tex]
and then we find b, which is the point where the curve intersects the y axis.
We can do it by plugging one point and the slope into the line equation form:
[tex]y=mx+b\\\\D(q)=mq+b\\\\138=\frac{-1}{25}(250) +b\\\\138=-10+b\\\\138+10=b=148[/tex]
With b: 148 and m: -1/25 we can write our demand equation D(q)
[tex]D(q)=\frac{-1}{25} q+148[/tex]
(b) to find the supply equation S(q) we have
P1: 102 Q1: 2200
P2: 102 Q2: 700
Similarly we find m, and b
[tex]m=\frac{p_{2} -p_{1} }{q_{2} -q_{1} }=\frac{72-102}{700-2200} =\frac{-30}{-1500}=\frac{1}{50}[/tex]
[tex]y=mx+b\\\\D(q)=mq+b\\\\72=\frac{1}{50}(700) +b\\\\72=14+b\\\\72-14=b=58\\[/tex]
And we can write our Supply equation S(q):
[tex]S(q)=\frac{1}{50}q+58[/tex]
(c) Now we may find the equilibrium quantity q* and the equilibrium price p* by writing D(q)=S(q), which means the demand equals the supply in equilibrium:
[tex]D(q)=S(q)\\\\\frac{-1}{25}q+148=\frac{1}{50}q+58\\\\[/tex]
[tex]148-58=\frac{1q}{50} +\frac{1q}{25} \\\\90= \frac{1q}{50} +\frac{2q}{50}\\\\90=\frac{3q}{50}\\ \\q=1500\\\\[/tex]
We plug 1500 as q into any equation, in this case S(q) and we get:
[tex]S(q)=\frac{1}{50}q+58\\\\S(1500)=\frac{1}{50}(1500)+58\\\\S(1500)=30+58\\\\S(1500)=88[/tex]
Which is the equilibrium price p*.
