Please help me on this Pure Maths question.

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ken74

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A company expects to sell 20000 computers in the first year if the price of each computer is £650. Let x represent the number of £’s by which the price has decreased.

a Write an expression for the price, p, of one computer, in the form p = a + bx. The company expects to sell an additional 50 computers every time the price decreases by £1.
b Write an expression for the number of computers sold, C, in the form C = d + ex. Revenue is defined by the formula, revenue = (number of computers sold) × (cost of one computer)
c Write an equation for revenue, r, in the form A – B(x – C) 2 , where A, B and C are constants to be found. The company wishes to maximise the revenue.
d Using your answer to part c, or othwerwise, state the price the company should charge for each computer and the revenue they will attain.
 

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This confusion here is the part a). I have no clue. How to do this please.
Write an expression for the price, p, of one computer, in the form p = a + bx. The company expects to sell an additional 50 computers every time the price decreases by £1.
 
ken74 said:
This confusion here is the part a). I have no clue. How to do this please.
Write an expression for the price, p, of one computer, in the form p = a + bx. The company expects to sell an additional 50 computers every time the price decreases by £1.
Click to expand...
Try to plot the price ' p' (y- axis) vs. the # of computers sold 'x' (x-axis). You are given:

p = a + b * x ..,..this is equation of a straight line

What does 'b' represent for the p-x line. You have been given the value of 'b'.

You have also given the co-ordinates of a point on the line. What is it?

Draw the line showing the axes and label the point ..... and continue.......
 
Hi Ken. The price starts at 650 pounds. If that price decreases by x pounds, then what expression could we write for the new price:

p = ?

The number of computers sold starts at 20000. If that quantity increases by 50 for each x, then what expression could we write for the new quantity:

c = ?

Your exercise should not use an upper-case C to represent two different numbers (a variable and a parameter), so I changed to a lower-case c for the number of computers sold.

:)

[imath]\;[/imath]
 
Otis said:
Hi Ken. The price starts at 650 pounds. If that price decreases by x pounds, then what expression could we write for the new price:

p = ?

The number of computers sold starts at 20000. If that quantity increases by 50 for each x, then what expression could we write for the new quantity:

c = ?

Your exercise should not use an upper-case C to represent two different numbers (a variable and a parameter), so I changed to a lower-case c for the number of computers sold.
Click to expand...
would p= a+b (650-x)
 
Subhotosh Khan said:
Try to plot the price ' p' (y- axis) vs. the # of computers sold 'x' (x-axis). You are given:

p = a + b * x ..,..this is equation of a straight line

What does 'b' represent for the p-x line. You have been given the value of 'b'.

You have also given the co-ordinates of a point on the line. What is it?

Draw the line showing the axes and label the point ..... and continue.......
Click to expand...

b is the slope which is a fixed value. Is it 650?. Im sorry Im still unclear
 
ken74 said:
would p= a+b [take this form:]
650-x
Click to expand...
Yes, except that you forgot to include the x in the definition for p. The variable price is:

p = a + b*x

Therefore,

p = 650 – x

where a=650 (the beginning price) and b=-1 (the number of pounds by which 650 changes for each x)

What expression represents the variable quantity of computers sold?

c = ?

[imath]\;[/imath]
 
Otis said:
Yes. The variable price is

p = 650 – x

That matches the form

p = a + bx

where a=650 and b=-1

What expression represents the variable quantity of computers sold?

c = ?

[imath]\;[/imath]
Click to expand...
C= 20000(650-x). Is this correct
 
ken74 said:
C= 20000(650-x). Is this correct
Click to expand...
No. The number of computers sold depends upon x, so c is not 20000.

Also, they are using symbol R for revenue, not C.

If the number of computers sold (c) starts at 20000 and increases by 50 for each x, then what expression could we write for c?

[imath]\;[/imath]
 
Otis said:
No. The number of computers sold depends upon x, so c is not a constant.

Also, they are using symbol R for revenue, not C.

If the number of computers sold starts at 20000 and increases by 50 for each x, what expression could we write for c?

[imath]\;[/imath]
Click to expand...
c= 20000 +x. Could this be right. since c is number of computer.
 
ken74 said:
c= 20000 +x
Click to expand...
That's better, but it doesn't show that 20000 increases by 50 for each x.

What you've written shows that c increases by 1 for each x.

Please try again.

[imath]\;[/imath]
 
Otis said:
That's better, but it doesn't show that 20000 increases by 50 for each x.

What you've written shows that c increases by 1 for each x.

Please try again.

[imath]\;[/imath]
Click to expand...
c=20000+ex. e is number of computer? I'm sorry still cant compute.
 
ken74 said:
x is price
Click to expand...
No, x is not price.

p is price.

x is the number of price decreases.

p = 650 – x

When there is one price reduction, then the price is 649 pounds.

p = 650 – 1 = 649

When there are six reductions, then the price is 644 pounds.

p = 650 – 6 = 644

[imath]\;[/imath]
ken74 said:
c = 20000 + ex
Click to expand...
That is the correct form, but what is e?

e is the increase in number of computers sold for each price reduction.

If there is one price reduction, then c is 20050.

If there are two price reductions then c is 20100.

If there are three prices reductions, then c is 20150.

c increases by 50 for each x.

In other words,

c = 20000 + ? * x

[imath]\;[/imath]
 
ken74 said:
c=20000+ex. e is number of computer? I'm sorry still cant compute.
Click to expand...
Otis said:
No, x is not price.

p is price.

x is the number of price decreases.

p = 650 – x

When there is one price reduction, then the price is 649 pounds.

p = 650 – 1 = 649

When there are six reductions, then the price is 644 pounds.

p = 650 – 6 = 644

[imath]\;[/imath]

That is the correct form, but what is e?

e is the increase in number of computers sold for each price reduction.

If there is one price reduction, then c is 20050.

If there are two price reductions then c is 20100.

If there are three prices reductions, then c is 20150.

c increases by 50 for each x.
Click to expand...
so c=d+ex. where d is 20000+ex, e changes from 0 to anything therefore c= 20000+change in e times x. am I correct
 
ken74 said:
so c=d+ex. where d is 20000+ex, e changes from 0 to anything therefore c= 20000+change in e times x. am I correct
Click to expand...
c Write an equation for revenue, r, in the form A – B(x – C)^2 , where A, B and C are constants to be found. The company wishes to maximise the revenue.

r=(d+ex)(650-x). Is this right for part c),

I mean ( 20000+ex) (650-x). How did they get ^2 from?
 
ken74 said:
d is 20000+ex, e changes from 0 to anything
Click to expand...
d is not 20000+ex

d is 20000

e is also a constant (it does not change).

Therefore, the number of computers sold is

c = 20000 + e*x

e is the amount by which 20000 increases for each x.

They gave you the value of e:

ken74 said:
The company expects to sell an additional 50 computers every time the price decreases by £1.
Click to expand...

c = 20000 + 50*x

That is how we show 20000 increasing by 50 for each x.

Just like p=650–x shows 650 decreasing by 1 for each x.

R = c * p

Can you now multiply the expressions for c and p to obtain an expression for revenue?

:)
 
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