- WE'VE BEEN STUDYING LINEAR EQUATIONS NOW
FOR A COUPLE OF WEEKS.
I WANTED TO TAKE A FEW MINUTES OUT OF YOUR TIME NOW
AND DISCUSS A BUSINESS APPLICATION
OF LINEAR EQUATIONS.
OUR GOAL TODAY IS TO SOLVE THESE PROBLEMS
USING LINEAR EQUATIONS.
RIGS LTD., A CLOTHING FIRM,
HAS FIXED COST OF $10,000 A YEAR.
THESE COSTS SUCH AS RENT, MAINTENANCE AND SO ON
MUST BE PAID NO MATTER HOW MUCH THE COMPANY PRODUCES.
TO PRODUCE X UNITS OF A CERTAIN KIND OF SUIT
WOULD COST $40 PER SUIT OR UNIT
IN ADDITION TO THE FIXED COST THAT IS THE VARIABLE COST
OF PRODUCING X OF THESE SUITS IS 20 X DOLLARS.
THESE COSTS ARE DUE TO THE AMOUNT PRODUCED
AND STEM FROM ITEMS
SUCH AS MATERIAL, WAGES, FUEL, AND SO ON.
THE TOTAL COST, C OF X, OF PRODUCING X UNITS A YEAR
IS GIVEN BY THE FUNCTION C.
C OF X IS = TO THE VARIABLE COST WHICH WAS GIVEN
AS 20 X PLUS THE FIXED COST = TO $10,000.
SO OUR COST FUNCTION IS = TO 20 X + 10,000.
GRAPH THE COST FUNCTION
AND THEN WHAT IS THE TOTAL COST OF PRODUCING 100 SUITS
AND 400 SUITS.
LET'S TAKE A LOOK AT THE GRAPH.
OF COURSE, THIS SHOULD REMIND YOU OF SLOPED INTERCEPT FORM
OF A Y WHERE OUR SLOPE IS = TO 20
AND OUR Y INTERCEPT IS = TO 10,000.
NOW, I ALREADY GRAPHED IT FOR YOU--
MAYBE A LITTLE DIFFICULT TO SEE, BUT X AXIS GOES BY 100'S
AND THIS WOULD BE THE NUMBER OF SUITS.
AND THE Y AXIS IS THE TOTAL COST IN DOLLARS
AND IT GOES BY THOUSANDS.
SO RIGHT HERE IS OUR Y INTERCEPT OF 10,000
AND SO FORTH.
AND YOU CAN FIND VARIOUS POINTS IN THE Y,
BUT I'VE ALREADY GRAPHED IT FOR US.
OKAY, THE QUESTION WAS HOW MUCH DOES IT COST
TO PRODUCE 100 SUITS?
WELL, BY LOOKING AT THE GRAPH
OF COURSE WE COULD JUST GO TO 100 UP TO THE FUNCTION
OVER TO THE Y AXIS
AND SEE THAT IT'S GOING TO BE $12,000.
400 SUITS OVER HERE ON THE FAR RIGHT GO UP TO THE FUNCTION
AND YOU CAN SEE THAT IT'S GOING TO BE
IT LOOKS LIKE $18,000.
NOW, LET'S GO BACK TO THE PREVIOUS SLIDE
JUST FOR A MINUTE.
ANOTHER WAY TO FIGURE THAT OUT
OF COURSE WOULD JUST BE TO FIND C AND 100.
JUST TO VERIFY OUR GRAPH IT WOULD BE 20 x 100 + 10,000
AND OF COURSE YOU CAN CHECK THIS,
BUT IT IS GOING TO BE $12,000.
AND FOR C AND 400 SAME PROCESS--
IT WOULD BE 20 x 400 NOW + THE FIXED COST OF 10,000
WHICH THIS IS GOING TO BE 8,000 + 10,000 OR 18,000
THE SAME AS WHAT THE GRAPH SHOWED.
OKAY AND I'M JUST DOING THIS SUBSTITUTION
INTO THE COST FUNCTION.
OKAY, LETS MOVE ALONG NOW.
WHEN THE BUSINESS SELLS AN ITEM
IT RECEIVES THE PRICE PAID BY THE CUSTOMER.
THIS IS NORMALLY GREATER THAN THE COST TO THE BUSINESS
FOR PRODUCING THE ITEM OR LETS HOPE SO.
THE TOTAL REVENUE THAT A BUSINESS RECEIVES
IS THE PRODUCT OF A NUMBER OF ITEMS SOLD
AND THE PRICE PAID PER ITEM.
SO IF RIGS LIMITED SELLS X SUITS AT $80 PER UNIT
OR PER SUIT,
THE TOTAL REVENUE WOULD SIMPLY BE 80 x X.
SO OUR REVENUE EQUATION OR REVENUE FUNCTION R OF X
IS = TO THE UNIT PRICE,
WHICH IS $80 x THE UNIT SOLD IS X--WHICH GIVES US 80 X.
REMEMBER THE COST OF COURSE IS THE MONEY GOING OUT THE DOOR,
REVENUE IS THE MONEY COMING IN THE DOOR.
THE TOTAL PROFIT,
WHICH IS WHAT MOST BUSINESSES ARE CONCERNED ABOUT,
THE TOTAL PROFIT THAT A BUSINESS RECEIVES
IS THE AMOUNT LEFT OVER
AFTER ALL THE COST HAVE BEEN SUBTRACTED FROM THE REVENUE.
THUS IF P OF X REPRESENTS THE TOTAL PROFIT
WHERE X ITEMS ARE PRODUCED AND SOLD,
WE HAVE THE PROFIT P OF X
IS = TO THE REVENUE R OF X - COST C OF X.
DETERMINE THE PROFIT AND DRAW IT'S GRAPH
USING THE SAME SET OF AXIS
AS WAS USED FOR THE GRAPH IN THE PREVIOUS EXAMPLE.
I'VE ALREADY GRAPHED THIS
SO LET'S GO AHEAD AND TAKE A LOOK AT IT.
HERE IN BLACK WE HAVE OUR COST EQUATION.
AND NOTICE AS YOU PRODUCE MORE AND MORE SUITS
THE COST GOES UP.
IN BLUE WE HAVE THE REVENUE EQUATION.
AGAIN NOTICE THAT AS YOU SELL MORE AND MORE UNITS
OF COURSE YOUR REVENUE IS GOING TO INCREASE.
NOW, WHEN YOUR COSTS ARE MORE THEN YOUR REVENUE
OF COURSE IN THIS REGION YOU HAVE A LOSS.
WHEN THE REVENUE IS HIGHER THEN THE COST
OF COURSE YOU HAVE A GAIN OR A PROFIT.
AND RIGHT HERE IN THE MIDDLE HERE
WE CALL THIS WHERE THERE IS NO PROFIT OR NO LOSS,
THIS WOULD BE THE BREAK EVEN POINT.
A COMPANY WILL BREAK EVEN AT THAT VALUE OF X
FROM WHICH P OF X = X,
MEANING YOU'RE NOT LOSING ANY MONEY
BUT YOU'RE NOT GAINING ANY MONEY EITHER.
THIS POINT IS AT WHICH THE REVENUE IS = TO THE COST.
FIND THE BREAK EVEN VALUE OF X.
REMEMBER THAT OUR REVENUE FUNCTION WAS 80 X
AND OUR COST FUNCTION WAS 20 X + 10,000.
SO IF WE WANT TO KNOW WHERE THAT POINT IS
WHERE WE WONT MAKE ANY MONEY, WE WONT LOOSE ANY MONEY,
WE NEED TO SET THESE TWO FUNCTIONS = TO EACH OTHER
AND SOLVE.
LET'S TRY IT.
WHEN IS THE REVENUE 80 X = TO THE TOTAL COST?
AND NOTICE THAT THE ONLY UNKNOWN IS THE NUMBER OF SUITS
THAT SOLVE FOR X.
SO I'LL SUBTRACT 20 ON BOTH SIDES--
SO I'LL SUBTRACT 20 X ON BOTH SIDES,
DIVIDE BY 60 ON BOTH SIDES.
X COMES OUT TO 166.6 REPEATING.
SO WE'D PROBABLY JUST ROUND THIS TO 167.
SO IN ORDER FOR THIS COMPANY TO BREAK EVEN SELLING SUITS
THEY'D HAVE TO SELL 167 SUITS.
NOW WITH THAT WE COULD GO BACK
AND WE COULD FIGURE OUT A LOT--
WITH THAT WE COULD GO BACK AND FIND SOME INITIAL INFORMATION
IF WE WANTED TO.
IF WE FOUND C OF 167 WE COULD FIND OUR TOTAL COST.
IF WE FOUND R OF 167 WE COULD GO BACK AND FIND
WHAT OUR REVENUE MUST BE TO BREAK EVEN.
AND THEN LASTLY, IF WE FIND P OF 167
SINCE THIS IS OUR BREAK EVEN POINT,
WE WOULD KNOW THAT SHOULD BE 0.
NOW THIS IS ROUNDED SO IT'LL BE A LITTLE BIT OFF,
BUT THAT SHOULD BE TRUE.
I HOPE THAT HELPS EXPLAIN HOW YOU CAN USE LINEAR EQUATIONS
TO SOLVE SOME BASIC BUSINESS APPLICATIONS.
HAVE A GREAT DAY.