Another application of mathematical function from a real-world situation is profit, revenue and cost functions.
Note that profit = revenue – cost.
P(x) = R(x) – C(x) where x= number of items sold.
Note: P(x) = positive value means it earns a profit and R(x) > C(x)
P(x) = negative value means loss and R(x) < C(x)
P(x) =0 or break-even where there is no loss or gain of profit. R(x)=C(x).
Suppose a store franchised a toy for $5 each with a fixed cost of $230.
Cost function = total variable cost + total fixed cost.
C(x) = 5x + 230.
The toys are sold for a price $8.
Revenue function = price * items
R(x) = 8*x or R(x) = 8x.
Profit function:
P(x) = 8x – (5x+230)
= 8x – 5x -230
= 3x -230
*Break-even when R(x) = C(x).
8x = 5x +230
-5x -5x
------------------------
3x = 0 +230
3x = 230
--- ------
3 3
x = 76.7
Round off to 77 toys since 0.70 toy is not possible.
The number of items can be counted as 0,1,2,3,…
Interpretation:
The store should at least had sold 77 toys to earn a profit.
When the sold items is less than 77 toys, there will be a loss.
Let x = 77 toys then profit is:
P(x) = 3(x) -230
P(77) = 3(77) -230
= 231 -230
= $1
The store earns $1 profit for 77 toys sold.
Function relates input values (independent values) to exactly one possible output value (dependent value).
Usually, it follows f(x) =y where x = input and y = output.
One common application is projectile motion: f(t) = `at^2 +bt+c`
where a <0 and f(t) as height of the object at time "t".
The graph of parabola opens downward (projectile path) when a <0.
Example: `f(t) = -4t^2 +14t+6 ` where f(t) is height in feet and t is time in seconds.
a) Finding initial height:
Let t=0 (starting time) then
f(0) =` -4(0)^2+14(0)+6` = 6 ft
Interpretation: An object is launched from the height of 6 feet.
b) Finding maximum height.
Hint: Vertex is at the peak of the parabola in projectile trajectory.
`Vertex t = -b/(2a) = -(14/(2*(-4))) =1.75 seconds`
maximum height: `f(3.5) = -4(3.5)^2 +14(3.5) +6 =18.25 feet.`
Interpretation: The object will reach 18.25 feet after 1.75 seconds before it falls down along its projectile trajectory.
c) Time it hits the ground.
At the ground level, the height is 0 or f(t)=0
f(t) = -4t^2 +14t +6
0 = -4t^2 +14t +6
0 = -(t-2)(4t+3) Factoring
Zero-factor property:
t-2 =0
+2 +2
---------------
t =2 seconds
4t +3 =0
-3 -3
----------------
4t = -3
--- ----
4 4
t= -0.75 seconds (negative time value does not exists).
Interpretation: The object hits the ground after 2 seconds.