# Math

• Post a relationship from a real-world situation or concept that illustrates the idea of a mathematical function. For example, we use the function F =9/5(C + 32) to convert from degrees Celsius to degrees Fahrenheit. 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.

Approved by eNotes Editorial Team 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.

Approved by eNotes Editorial Team