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 boxes and the demand is 300 boxes. (a... 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 boxes and the demand is 300 boxes. (a) Find a price-supply equation of the form P = mQs + b, where P is the price in dollars and Qs is the corresponding supply in thousands of boxes. (b) Find a price-demand equation of the form P = mQd + b, where P is the price in dollars and Qd is the corresponding demand in thousands of boxes. (c) Find the equilibrium point of supply and demand algebraically and graphically. 2. Given the supply and demand functions P = Qs^2 + 14Qs + 22, calculate the equilibrium price and quantity. Read more

Steps to Solve

1. Finding the Price-Supply Equation

   To find the price-supply equation of the form $P = m Q_s + b$, we can use the two points given:

   • Point 1: $(Q_s = 320, P = 9.00)$
   • Point 2: $(Q_s = 270, P = 8.50)$.

   We will first convert the supply values from boxes to thousands of boxes:

   • Point 1: $Q_s = 0.32$
   • Point 2: $Q_s = 0.27$.

   Next, we calculate the slope $m$ using the formula:

   $$ m = \frac{P_2 - P_1}{Q_{s2} - Q_{s1}} = \frac{8.50 - 9.00}{0.27 - 0.32} = \frac{-0.50}{-0.05} = 10. $$

   Now we can use point-slope form to find $b$:

   $$ P = m Q_s + b \implies 9 = 10(0.32) + b $$

   Solving for $b$ gives:

   $$ b = 9 - 3.2 = 5.8. $$

   Therefore, the price-supply equation is:

   $$ P = 10 Q_s + 5.8. $$

2. Finding the Price-Demand Equation

   To find the price-demand equation of the form $P = m Q_d + b$, we use the points given:

   • Point 1: $(Q_d = 200, P = 9.00)$
   • Point 2: $(Q_d = 300, P = 8.50)$.

   We convert demand values to thousands of boxes:

   • Point 1: $Q_d = 0.2$
   • Point 2: $Q_d = 0.3$.

   We calculate the slope $m$:

   $$ m = \frac{P_2 - P_1}{Q_{d2} - Q_{d1}} = \frac{8.50 - 9.00}{0.3 - 0.2} = \frac{-0.50}{0.1} = -5. $$

   Now use point-slope form to find $b$:

   $$ 9 = -5(0.2) + b \implies b = 9 + 1 = 10. $$

   The price-demand equation is:

   $$ P = -5 Q_d + 10. $$

3. Finding the Equilibrium Point Algebraically

   To find the equilibrium, set the supply equation equal to the demand equation:

   $$ 10 Q_s + 5.8 = -5 Q_d + 10. $$

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

   $$ 10 Q + 5.8 = -5 Q + 10. $$

   Rearranging gives:

   $$ 15 Q = 10 - 5.8 \implies 15 Q = 4.2 \implies Q = \frac{4.2}{15} = 0.28. $$

   Then, plug $Q$ back into one of the equations to find $P$:

   $$ P = 10(0.28) + 5.8 = 2.8 + 5.8 = 8.6. $$

   The equilibrium point is $(Q = 0.28 \text{ thousands of boxes}, P = 8.60).$

4. Finding the Equilibrium Point Graphically

   To visualize, plot the price-supply and price-demand equations on a graph with $P$ on the y-axis and $Q$ on the x-axis. The intersection point of these two lines will confirm the equilibrium point $(Q = 0.28, P = 8.60).$

5. Finding Equilibrium for the Given Functions

   For the second part, we have the functions:

   $$ P = Q_s^2 + 14Q_s + 22 $$

   To find equilibrium, set supply equal to demand. However, the demand function is not given explicitly. If we assume $Q_d = Q_s$, we can only utilize the provided function's coefficients in solving additional problems.

   We can rewrite it for potential point analysis as we equate:

   $$ P = Q_s^2 + 14Q_s + 22 = P_d. $$

   Graphing or simplifying further might give the intersection values.