Systems of linear equations can be used in various subject areas and are used to solve for unknowns. With n equations, the highest possible number of unknowns that can be solved for is n. Complications occur when the number of unknowns is unequal to the number of equations in the system.
Linear systems of equations with more or less equations than unknowns occur quite often. When there are fewer equations than unknowns, it is impossible to solve for all the unknowns algebraically. Because of this, there are usually infinitely many solutions for each variable (it is hard to narrow it down). When there are more equations than unknowns (an overdetermined system), there are usually no solutions. This occurs because there may be solutions that satisfy two of the three equations, but not necessarily all three. The only way there would be solutions to an overdetermined system (for example, one with three equations and two unknowns) would be if all the equations are equivalent, if two of the three equations are equivalent and the other equation intersects both, or if all three equations intersect each other at a single point.
Economics
I. Linear Equations can be used to find out the best method of maximizing profits. Let's say a farmer produces 2 goods, yogurt and ice cream. Each quart of yogurt requires 0.4 quarts of milk and 0.2 quarts of cream, while each quart of ice cream requires 0.2 quarts of milk and 0.4 quarts of cream. The profit gained from a quart of yogurt is 8 cents and the profit gained from a quart of ice cream is 10 cents. If the farmer has 10 quarts of milk and 14 quarts of cream, how much of each good should he make to maximize his profits?
Let y = number of quarts of yogurt and let i = number of quarts of ice-cream. We now have an equation for the total amount of milk required:
0.4y + 0.2i
The equation for cream would look like this:
0.2y + 0.4i
Since there are only 10 quarts of milk and 14 quarts of cream, we have to alter these equations slightly:
0.4y + 0.2i ≤ 10
0.2y + 0.4i ≤ 14
An equation for total profit can also be made, as we know how much profit is obtained from a quart of each good. Let p = profit.
p = 8y + 10i
We know that y and i cannot be negative, because there is no such thing as a negative quart value. Using the two inequalities we obtained, we can create a graph which will give us a region of points which are feasible. This region is a convex polygon, so we must test all the corner points (excluding the origin) to find the maximum profit to be obtained. These points are A(0,35), B(10,30) and C(25,0). The value of the objective function is largest at point B, so we will plug the values of point B into our equations.
P = 8(10) + 10(30)
P = $3.80
The farmer's optimal profit would be $3.80 a day, using his materials to create 10 quarts of yogurt and 30 quarts of ice-cream.
**Profit maximization is very important for an economist. This is an invaluable source of revenue for an economist as a good plan will make much more money than a bad one. It is important to use linear equations in order to find out the best way to make money.
II. Linear equations can be used to find the equilibrium price and quantity to be supplied in a given market. Let's say there are two products, orange juice and water, which are interrelated. Let p1 and q1 represent the price and quantity demanded respectively for product 1 (orange juice) and p2 and q2 represent the same for product 2 (water).
Demand Supply
Product 1: p1 = 2000 - 3q1 - 2q2 q1 = 100 + 2q1 + q2
Product 2: p2 = 2800 - q1 - 4q2 q2 = 200 + 3q1 + 2q2
For equilibrium to be achieved, both price expressions must be equal, so the following equations are obtained:
Product 1: 2000 - 3q1 - 2q2 = 100 + 2q1 + q2
Product 2: 2800 - q1 - 4q2 = 200 + 3q1 + 2q2
Simplifying these expressions gives us the equations
5q1 + 3q2 = 1900
4q1 + 6q2 = 2600
Elimination can now be utilized.
4(5q1 + 3q2 = 1900)
-
5(4q1 + 6q2 = 2600)
=
-18q2 = -5400, which is equal to q2 = 300. Plugging this value into the equation 4q1 + 6q2 = 2600 gives us 4q1 + 1800 = 2600, which can be simplified to q1 = 200.
By plugging these values (of the quantities) into the original demand equations, we find that the equilibrium price for product 1 is $800 and the equilibrium price for product 2 is $1400.
p1 = $800
q1 = 200 units
p2 = $1400
