Questions: At a price of 21, the supply is about 1680 watches.
(Round to the nearest whole number as needed.)
g. Graph p=S(q)=1.25 q on the same axis used to graph p=28-2.25 q in part d. Choose the correct graph below.
A.
B.
c.
D.
h. Given that the demand function is p=D(q)=28-2.25 q and that the supply function is p=S(q)=1.25 q, find the equilibrium quantity and the equilibrium price.
The equilibrium quantity is watches.
Transcript text: At a price of \$21, the supply is about 1680 watches.
(Round to the nearest whole number as needed.)
g. Graph $p=S(q)=1.25 q$ on the same axis used to graph $p=28-2.25 q$ in part d. Choose the correct graph below.
A.
B.
c.
D.
h. Given that the demand function is $\mathrm{p}=\mathrm{D}(\mathrm{q})=28-2.25 \mathrm{q}$ and that the supply function is $\mathrm{p}=\mathrm{S}(\mathrm{q})=1.25 \mathrm{q}$, find the equilibrium quantity and the equilibrium price.
The equilibrium quantity is $\square$ watches.
Solution
Solution Steps
Step 1: Find the intersection point of the two lines
The two lines represent the supply and demand functions:
Supply: \(p = 1.25q\)
Demand: \(p = 28 - 2.25q\)
The intersection point represents the equilibrium point where supply equals demand. We can find this by setting the two equations equal to each other:
\(1.25q = 28 - 2.25q\)
Step 2: Solve for q (equilibrium quantity)
Add \(2.25q\) to both sides of the equation:
\(1.25q + 2.25q = 28\)
\(3.5q = 28\)
Divide both sides by 3.5:
\(q = \frac{28}{3.5}\)
\(q = 8\)
Step 3: Solve for p (equilibrium price)
Substitute the value of \(q\) (8) back into either the supply or demand equation. Using the supply equation:
\(p = 1.25 * 8\)
\(p = 10\)
Step 4: Identify the correct graph
Graph A correctly shows the intersection of the two lines at q = 8. The demand line (red) starts at p=28 and decreases, while the supply line (blue) starts at the origin and increases. Their intersection point in graph A corresponds to q=8 and p=10.
Final Answer
The equilibrium quantity is \(\boxed{8}\) watches.
The equilibrium price is \(\boxed{10}\).
The correct graph is \(\boxed{A}\).