# Ex: Linear Equation Application (Write a Profit Equation) – HERE’S AN APPLICATION
PROBLEM INVOLVING REVENUE COSTS
AND PROFIT THAT CAN BE SOLVED
USING A LINEAR EQUATION. A COMPANY MAKES PHONE COVERS. THEY SELL EACH COVER FOR \$12. IF X EQUALS THE NUMBER
OF PHONE COVERS PRODUCED AND SOLD, WHAT IS THE REVENUE EQUATION? WELL THE REVENUE WOULD BE
THE TOTAL AMOUNT OF MONEY GENERATED FROM THE SALE
OF THE PHONE COVERS SO WE CAN SAY
THAT THE REVENUE BIG R, IS GOING TO BE EQUAL TO 12 x X WHERE AGAIN X IS THE NUMBER
OF PHONE COVERS PRODUCED AND SOLD. AND R WOULD BE
THE TOTAL REVENUE IN DOLLARS. NEXT THE COST TO MAKE
THE COVERS IS \$2 EACH WITH A ONE TIME START UP COST
OF \$5,200. SO NOW WE WANT TO KNOW
WHAT THE COST EQUATION IS. AND WE CAN SEE THERE ARE
TWO COMPONENTS TO THE COST, THERE’S THIS FIXED COST
OF \$5,200 THAT’S NOT GOING TO CHANGE AND THEN WE HAVE
A VARIABLE COST BASED UPON THE NUMBER OF PHONE
COVERS PRODUCED AT \$2 EACH. SO THE TOTAL COST IS GOING TO BE EQUAL TO 2 x THE NUMBER
OF PHONE COVERS PRODUCED, PLUS THE FIXED COSTS,
OR START UP COSTS OF \$5,200. SO C IS GOING TO REPRESENT
THE TOTAL COST IN DOLLARS WHERE X IS THE NUMBER
OF PHONE COVERS PRODUCED. AND IT’S PRETTY TYPICAL
TO LEAVE THE UNITS OFF IN THE EQUATION AND JUST MAKE A NOTE FOR
WHAT THE VARIABLES REPRESENT. NOW WE WANT TO DETERMINE
THE PROFIT WHEN THE COMPANY SELLS
1,200 COVERS. SO THE FIRST THING
WE SHOULD DO IS WRITE THE PROFIT EQUATION AND PROFIT IS EQUAL
TO REVENUE MINUS COSTS. AGAIN WHERE REVENUE IS
THE AMOUNT OF MONEY COMING IN AND THE TOTAL COST IS
THE AMOUNT OF MONEY GOING OUT. SO THE PROFIT, BIG P, IS GOING TO BE EQUAL
TO THE REVENUE EQUATION OF 12X – THE COST EQUATION
OF 2X + 5,200. NOW IT IS IMPORTANT
THAT WE ADD PARENTHESIS HERE BECAUSE WE DO HAVE TO SUBTRACT
THE ENTIRE COST EQUATION. WE DO HAVE LIKE TERMS HERE
AND HERE SO WE CAN SIMPLIFY
THIS EQUATION. WE’RE GOING TO HAVE
THE PROFIT, IT’S GOING TO BE EQUAL
TO 12X – 2X, THAT WOULD BE 10X AND THEN THIS WOULD BE
MINUS 5,200. SO NOW WE CAN USE
THIS EQUATION TO DETERMINE THE PROFIT WHEN THE COMPANY SELLS
1,200 COVERS. THE PROFIT WOULD BE EQUAL
TO 10 x 1,200 – 5,200. WELL HERE WE’D HAVE
12,000 – 5,200 SO THIS DIFFERENCE WOULD BE
THE COMPANY’S PROFIT FROM SELLING 1,200 COVERS SO THE PROFIT IS GOING TO BE
EQUAL TO– THIS WOULD BE \$6,800. AND THE LAST QUESTION ASKS US
TO FIND THE NUMBER OF COVERS THEY NEED TO SELL
TO BREAK EVEN. WELL BREAK EVEN IS WHEN THE
COMPANY DOESN’T MAKE ANY MONEY OR LOSE ANY MONEY. SO THAT WOULD OCCUR WHEN THE REVENUE
IS EXACTLY EQUAL TO THE COST. SO TO DETERMINE
THE BREAK EVEN POINT WE’LL HAVE TO SET THE REVENUE
EQUAL TO THE COST WHICH MEANS WE’LL HAVE
12X=TO 2X + 5,200. SO TO SOLVE FOR X HERE, WE’LL
SUBTRACT 2X ON BOTH SIDES. WE WOULD HAVE 10X=5,200. LET’S GO AHEAD
AND FINISH THIS UP HERE. SO WE’LL DIVIDE BOTH SIDES
BY 10, SO WE HAVE X=5,200 DIVIDED
BY 10 WHICH WOULD BE 520. SO THEY NEED TO SELL 520
COVERS TO BREAK EVEN, MEANING THEY WON’T MAKE
ANY MONEY OR LOSE ANY MONEY. I HOPE YOU FOUND THIS HELPFUL.

### One thought on “Ex: Linear Equation Application (Write a Profit Equation)”

• September 30, 2017 at 4:28 pm