1. At a price of $9.00 per box of oranges, the supply is 320 boxes and the demand is 200 boxes. At a price of $8.50 per box, the supply is 270 and the demand is 300 boxes. (a) Find... 1. At a price of $9.00 per box of oranges, the supply is 320 boxes and the demand is 200 boxes. At a price of $8.50 per box, the supply is 270 and the demand is 300 boxes. (a) Find a price-supply equation of the form P = mQs + b. (b) Find a price-demand equation of the form P = mQd + b. (c) Find the equilibrium point of supply and demand algebraically and graphically. Read more

Steps to Solve

1. Finding the Supply Equation

We start with the given supply data. We know that at a price of $9, the supply is 320 boxes ($Q_s = 320$) and at a price of $8.50, the supply is 270 boxes ($Q_s = 270$).

We can represent the supply function in the form $P = m Q_s + b$. To find the parameters, we'll calculate the slope $m$:

$$ m = \frac{P_2 - P_1}{Q_{s2} - Q_{s1}} = \frac{8.50 - 9.00}{270 - 320} = \frac{-0.50}{-50} = 0.01 $$

Now we can use one of the points to solve for $b$. Using the point $(320, 9)$:

$$ 9 = 0.01 \cdot 320 + b $$

$$ b = 9 - 3.2 = 5.8 $$

Thus, the supply equation is:

$$ P = 0.01 Q_s + 5.8 $$

2. Finding the Demand Equation

Next, we will set up the demand function using the given data. At a price of $9.00, the demand is 200 boxes ($Q_d = 200$) and at a price of $8.50, the demand is 300 boxes ($Q_d = 300$).

Similarly, we calculate the slope $m$ of the demand function:

$$ m = \frac{P_2 - P_1}{Q_{d2} - Q_{d1}} = \frac{8.50 - 9.00}{300 - 200} = \frac{-0.50}{100} = -0.005 $$

Using one of the points, say $(200, 9)$, we solve for $b$ in $P = m Q_d + b$:

$$ 9 = -0.005 \cdot 200 + b $$

$$ b = 9 + 1 = 10 $$

Thus, the demand equation is:

$$ P = -0.005 Q_d + 10 $$

3. Finding the Equilibrium Point

To find the equilibrium quantity, we set the supply and demand equations equal to each other:

$$ 0.01 Q_s + 5.8 = -0.005 Q_d + 10 $$

Since at equilibrium $Q_s = Q_d = Q$, we substitute $Q$:

$$ 0.01 Q + 5.8 = -0.005 Q + 10 $$

Combining the terms gives:

$$ 0.015 Q = 4.2 $$

Solving for $Q$ yields:

$$ Q = \frac{4.2}{0.015} = 280 $$

Now we can substitute $Q$ back into either the supply or demand function to find the equilibrium price. Using the supply function:

$$ P = 0.01 \cdot 280 + 5.8 = 2.8 + 5.8 = 8.6 $$

Thus, the equilibrium point is at $P = 8.60$ and $Q = 280$ boxes.