given p=68-4qd-qd^2
p=12+2qs+qs^2
where p is the price
qd is the quantity demanded
qs is the quantity supplied
determine the uquilibrium and the quantity of the commodity
awesome
I request for a solution
To find the equilibrium point in this scenario, we need to find the values of qd (quantity demanded) and qs (quantity supplied) that satisfy both equations simultaneously.
Step 1: Set the equations equal to each other.
Since p is the same on both sides of the equation, we can set the right sides of the equations equal to each other:
68 - 4qd - qd^2 = 12 + 2qs + qs^2
Step 2: Rearrange the equation to have the same variable on one side.
Rearrange the equation by combining like terms and moving all the terms to one side:
qd^2 + qs^2 + 4qd - 2qs - 56 = 0
Step 3: Solve the quadratic equation.
Since we have a quadratic equation in terms of qd and qs, we can use the quadratic formula or factor to solve for qd and qs. Let's use factoring to simplify the equation:
(qd + 7)(qd - 4) + (qs + 7)(qs - 2) = 0
Now, we have two separate equations:
qd + 7 = 0 ---> qd = -7
qs + 7 = 0 ---> qs = -7
qd - 4 = 0 ---> qd = 4
qs - 2 = 0 ---> qs = 2
So, we have two possible sets of values for (qd, qs): (-7, -7) and (4, 2).
Step 4: Substitute the values of qd and qs into either equation to find the corresponding price (p).
Let's substitute the values of qd and qs into the first equation:
p = 68 - 4qd - qd^2
p = 68 - 4(-7) - (-7)^2
p = 92
The price at equilibrium when qd = -7 and qs = -7 is 92.
Similarly, if we substitute the values of qd and qs into the second equation:
p = 12 + 2qs + qs^2
p = 12 + 2(2) + (2)^2
p = 20
The price at equilibrium when qd = 4 and qs = 2 is 20.
Therefore, there are two equilibrium points:
1. (qd, qs) = (-7, -7)
p = 92
2. (qd, qs) = (4, 2)
p = 20
These equilibrium points represent the quantity of the commodity demanded and supplied at their corresponding prices.