Ex: Linear Inequality in Two Variables Application Problem (Phone Cost: Day and Night Minutes)


– A PHONE COMPANY CHARGES
20 CENTS FOR DAYTIME MINUTES AND FIVE CENTS
FOR NIGHTTIME MINUTES. HOW MANY DAY AND NIGHT MINUTES
WOULD YOU HAVE TO USE TO BE CHARGED MORE THAN
10 DOLLARS IN 24 HOURS? SO THIS SITUATION REPRESENTS
AN INEQUALITY IN TWO VARIABLES. SO LET’S LET D EQUAL THE NUMBER
OF DAY MINUTES, AND LET’S LET N EQUAL
THE NUMBER OF NIGHT MINUTES. AND THERE’S ONE MORE ISSUE
TO DEAL WITH. THE CHARGE PER MINUTE HERE
IS IN CENTS, BUT THEN WE WANT TO ANSWER
THE QUESTION BASED UPON BEING CHARGED
MORE THAN 10 DOLLARS. SO WE HAVE TO EITHER CONVERT
THE CENTS TO DOLLARS OR DOLLARS TO CENTS. AND TO AVOID DECIMALS, LET’S
CONVERT 10 DOLLARS TO CENTS. SINCE ONE DOLLAR=100 CENTS, THEN 10 DOLLARS WOULD BE EQUAL
TO 10 x 100 OR 1,000 CENTS. NOW, LET’S SEE IF WE CAN WRITE
AN INEQUALITY TO REPRESENT THIS SITUATION. EVERY DAYTIME MINUTE COSTS
20 CENTS. SO THE AMOUNT OF MONEY SPENT
ON DAYTIME MINUTES WOULD BE 20 CENTS X D + THE AMOUNT OF MONEY SPENT
ON NIGHTTIME MINUTES WOULD BE FIVE CENTS x THE NUMBER
OF NIGHT MINUTES OR 5 x N. WE WANT TO KNOW WHEN THIS WOULD
BE MORE THAN 10 DOLLARS IN 24 HOURS OR>1,000 CENTS. SO NOW, WE’LL GO AHEAD
AND GRAPH THIS INEQUALITY WHERE THE GRAPH WOULD SHOW
ALL THE POSSIBLE SOLUTIONS TO THIS GIVEN INEQUALITY. SO WE’LL LET THE HORIZONTAL AXIS
EQUAL THE D AXIS, WHICH REPRESENTS THE NUMBER
OF DAY MINUTES. AND WE’LL LET THE VERTICAL AXIS
BE THE N AXIS, WHICH REPRESENTS THE NUMBER
OF NIGHT MINUTES. SO WE NEED TO GRAPH
THIS INEQUALITY. SO THE FIRST STEP IS TO GRAPH
THE BORDERLINE. SO WE’LL GO AHEAD AND GRAPH
THE LINE 20D + 5N=1,000, AND BECAUSE THE ORIGINAL
INEQUALITY IS>, THE LINE WILL NOT BE PART
OF THE SOLUTION, AND TO SHOW THIS,
WE’LL MAKE A DASHED LINE. AND SINCE THIS LINE
IS IN STANDARD FORM, LET’S GO AHEAD AND FIND THE X
AND Y INTERCEPTS TO GRAPH THE LINE OR IN THIS CASE,
THE D AND N INTERCEPT. SO TO FIND THE D INTERCEPT,
WE’LL SET N=0, AND THEN TO FIND
THE N INTERCEPT, WE’LL SET D=0. SO IF WE SET N=0,
5N WOULD BE 0, AND WE’D HAVE 20D=1,000. DIVIDE BOTH SIDES BY 20,
AND WE HAVE D=50. SO THE D INTERCEPT=50,
AND NOW WE’LL SET D=0. AND IF D=0, 20D WOULD BE 0,
AND WE’D HAVE 5N=1,000. DIVIDE BOTH SIDES BY 5,
AND WE’D HAVE N=200. SO NOW WE CAN PLOT THESE
TWO POINTS AND SKETCH OUR LINE, BUT BEFORE WE DO THIS, LET’S GO
AHEAD AND SCALE OUR AXES. WE CALL THIS 200,
AND THIS WOULD BE 100. IF WE LET THIS EQUAL 50,
THEN THIS WOULD BE 25. SO NOW,
WE’LL PLOT THE POINT (50, 0), WHICH IS THE D INTERCEPT,
THIS POINT HERE. AND NOW, WE’LL PLOT THE POINT
(0, 200), WHICH IS THE N INTERCEPT,
WHICH IS THIS POINT HERE. AND NOW, WE’LL SKETCH A LINE
THROUGH THOSE TWO POINTS, AND AGAIN, WE SAID
IT WAS GOING TO BE DASHED BECAUSE OF THE VISUAL INEQUALITY
SYMBOL. SO IT’S GOING TO LOOK SOMETHING
LIKE THIS, AND NOW, WE NEED TO FIGURE OUT WHETHER WE’RE GOING TO
SHADE BELOW THE LINE OR ABOVE THE LINE. AND NOTICE HOW WE’RE ONLY
CONSIDERING THE FIRST QUADRANT, BECAUSE WE KNOW THAT
THE NUMBER OF MINUTES CAN NEVER BE NEGATIVE. NOW THAT WE HAVE THE GRAPH
OF OUR BORDERLINE, WE HAVE TO FIGURE OUT
WHETHER WE SHOULD SHADE ABOVE OR BELOW THE LINE. SO IF WE SELECT THIS POINT HERE
AS OUR TEST POINT WITH THE COORDINATES (0, 0), IT’S GOING TO BE VERY EASY TO
SEE THAT WHEN D=0 AND N=0, 0 IS NOT>1,000, AND THEREFORE, THIS POINT DOES
NOT SATISFY THE INEQUALITY. SO WE HAVE TO SHADE
THE OTHER SIDE OF THE LINE OR THIS SIDE HERE. SO ANY POINT IN THE SHADED
REGION REPRESENTS A COMBINATION OF THE NUMBER OF DAY AND NIGHT
MINUTES THAT WOULD HAVE A CHARGE
OF MORE THAN 10 DOLLARS. SO WE COULD PROBABLY FIGURE OUT
LOGICALLY THAT IF THIS WAS THE LINE WHERE
THE CHARGES WOULD BE EXACTLY 10 DOLLARS OR 1,000 CENTS, THEN ANY POINT ABOVE THIS LINE WOULD REPRESENT THE COMBINATION
OF DAY AND NIGHT MINUTES THAT WOULD BE MORE THAN
10 DOLLARS, AND THEREFORE,
WOULD BE IN THE REGION THAT WOULD REPRESENT
THE SOLUTION TO THIS LINEAR INEQUALITY. I HOPE THIS EXAMPLE HELPS.  

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