Math Is Fun Forum

mathland

Member

Registered: 2021-03-25

Posts: 444

Maximum Profit

The cost per unit in the production
of an MP3 player is $60. The manufacturer charges $90
per unit for orders of 100 or less. To encourage large
orders, the manufacturer reduces the charge by $0.15
per MP3 player for each unit ordered in excess of 100
(for example, the charge is reduced to $87 per MP3
player for an order size of 120).

(a) The table shows the profits P (in dollars) for
various numbers of units ordered, x. Use the table
to estimate the maximum profit.

Units x:         Profit P

130................3315   
140................3360   
150................3375
160................3360
170................3315

When I look at this table and think of the word MAXIMUM, I quickly think of the largest Profit P number. If that is the case, I say the maximum profit is 3375. Yes?

(b) Plot the points (x, P) from the table in part (a). Does
the relation defined by the ordered pairs represent P
as a function of x?

The same value of x cannot be matched to 2 different P-values (in this example).
For this reason, I say the ordered pairs represent P as a function of x. Part (c) below confirms that P is a function of x.

(c) Given that P is a function of x, write the function
and determine its domain. (Note: P = R − C,
where R is revenue and C is cost.)

I need help with part (c).

Bob

Administrator

Registered: 2010-06-20

Posts: 10,619

Re: Maximum Profit

I'm trying to build up an equation from the information.

P = R - C

C is fixed.  R depends on number sold ; let's call that x.

The discount only clicks in above an order of 100, so the term needed is (x-100)

So build up an R value using the percent reduction and (x-100)

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

mathland

Member

Registered: 2021-03-25

Posts: 444

Re: Maximum Profit

I'm trying to build up an equation from the information.

P = R - C

C is fixed.  R depends on number sold ; let's call that x.

The discount only clicks in above an order of 100, so the term needed is (x-100)

So build up an R value using the percent reduction and (x-100)

Bob

You said:

"...build up an R value using the percent reduction and (x-100)."

What do you mean?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,619

Re: Maximum Profit

Let's work through some number cases first.  Usually I find the algebra gets clearer when I do that.

x = 50

Cost = 50 x 60

Revenue = 50 x 90

Profit = R - C = 50 x(90-60)

I can see that for x less than 100 this is straight forward

R = 30x

What happens at exactly x = 100?  Do we have a discount or not?

in excess of 100

So no discount for exactly 100.

Above 100.

It has taken me 30 minutes to make sense of this.  I used a spreadsheet to backtrack the given profits to work out what the sale cost of a unit is for 120, 130, 140, 150, 160, 170.  The wording doesn't make this clear at all.

Here's my attempt at clarity.

If x > 100, calculate how much above 100 it is.  ie. calculate x -100

Multiply this by 0.15 to get the discount                       d =  (x-100) times 0.15

Subtract this from 90 to get the lower price of a unit     p = 90 - d

Work out the profit from one unit sold                          pr = p - 60

Finally multiply by the x value to get the revenue           R = pr times  x

Here's a couple of calculations to show this in  practice:

x = 130          d  =  30 times 0.15  = 4.5      p = 90 - d = 85.5      pr  = 85.5 - 60 = 25.5     R = 25.5 times 130 = 3315

x = 170          d  =  70 times 0.15  = 10.5    p = 90 - d = 79.5      pr  = 79.5 - 60 = 19.5     R = 19.5 times 170 = 3315

Now your turn.  Put the above algorithm together as a single formula in x

Best of luck,

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

mathland

Member

Registered: 2021-03-25

Posts: 444

Re: Maximum Profit

Let's work through some number cases first.  Usually I find the algebra gets clearer when I do that.

x = 50

Cost = 50 x 60

Revenue = 50 x 90

Profit = R - C = 50 x(90-60)

I can see that for x less than 100 this is straight forward

R = 30x

What happens at exactly x = 100?  Do we have a discount or not?

in excess of 100

So no discount for exactly 100.

Above 100.

It has taken me 30 minutes to make sense of this.  I used a spreadsheet to backtrack the given profits to work out what the sale cost of a unit is for 120, 130, 140, 150, 160, 170.  The wording doesn't make this clear at all.

Here's my attempt at clarity.

If x > 100, calculate how much above 100 it is.  ie. calculate x -100

Multiply this by 0.15 to get the discount                       d =  (x-100) times 0.15

Subtract this from 90 to get the lower price of a unit     p = 90 - d

Work out the profit from one unit sold                          pr = p - 60

Finally multiply by the x value to get the revenue           R = pr times  x

Here's a couple of calculations to show this in  practice:

x = 130          d  =  30 times 0.15  = 4.5      p = 90 - d = 85.5      pr  = 85.5 - 60 = 25.5     R = 25.5 times 130 = 3315

x = 170          d  =  70 times 0.15  = 10.5    p = 90 - d = 79.5      pr  = 79.5 - 60 = 19.5     R = 19.5 times 170 = 3315

Now your turn.  Put the above algorithm together as a single formula in x

Best of luck,

Bob

I will work on this a bit more.