solving linear equation

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solving linear equation

Postby Bronwyn » Thu Feb 18, 2021 7:56 am

I'm confused about how write linear equations to find the answers to the following:

Cost to make 3 items is \$125, cost to make 7 is \$125.

A. Assume liner relationship, determine rule for the cost C of making products s

Product sells for $25

B. Rule for the revenue R received from selling product s

C. Profit = revenue -cost. Rule for profit from selling products s. How many must be sold to break even

D. If she sells an average if t, Min price required to sell product s to breakeven. What if t =5


Any help or formulas would be greatly appreciated.

Thank-you
Bronwyn
 
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Re: solving linear equation

Postby Guest » Fri Mar 12, 2021 8:36 pm

Do you understand that a "linear function" is of the form y= ax+ b where x and y are the variables and a and b are constants? To "find a linear function" or "determine a linear function" means to find or determine values for a and b. To find two values, typically you need two equations.

"Cost to make 3 items is 125, cost to make 7 is 125, cost to make 7 is 125."
So when "x" is 3, "y" is 125 so 125= 3a+ b.
And when "x" is 7, "y'' is 175 so 125= 7a+ b. Subtract to get 0= 4a so a= 0. Then both equations reduce to b= 125. So the cost to make any number is 125! That should have been clear when both 3 and 7 were 125.
(I seem to remember doing this problem recently!)

The 'revenue" is how much money you bring in. If you sell each one for 25 then your revenue on selling "s" of them will be 25s. The profit will be 25s- 125. That will be negative if s is less than 5, positive if s is larger than 5. s= 5, when the profit is 0, is the "break even" point.
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Re: solving linear equation

Postby Guest » Sun Mar 28, 2021 12:30 pm

A silly typo:To "find a linear function" or "determine a linear function" means to find or determine values for a and b. To find two values, typically you need two equations.

"Cost to make 3 items i
Do you understand that a "linear function" is of the form y= ax+ b where x and y are the variables and a and b are constants? s 125, cost to make 7 is 125, cost to make 7 is 125."
So when "x" is 3, "y" is 125 so 125= 3a+ b.btra
And when "x" is 7, "y'' is 175 so 125= 7a+ b.

No, 175= 7a+ b
Subtract to get 0= 4a so a= 0. Then both equations reduce to b= 125. So the cost to make any number is 125! That should have been clear when both 3 and 7 were 125.
(I seem to remember doing this problem recently!)

Subtracting gives 50= 4a so a= 12,5. Then 125= 5(12.5)+ b= 62.5+ b so b= 62.5. y= 12.5x+ 62.5

The 'revenue" is how much money you bring in. If you sell each one for 25 then your revenue on selling "s" of them will be 25s. The profit will be 25s- 125. That will be negative if s is less than 5, positive if s is larger than 5. s= 5, when the profit is 0, is the "break even" point.
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Re: solving linear equation

Postby Guest » Wed Apr 14, 2021 11:56 am

I recommend you go back and read the problem again!
You wrote "Cost to make 3 items is $125, cost to make 7 is $125."

So the cost to make 7 is the same as the cost to make 3. That would be very strange and would make the problem trivial. If that is true, and the relationship is linear, then it must be a horizontal straight line- the cost to make any number would be $125.
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Re: solving linear equation

Postby stephenwoodford » Wed Jul 21, 2021 2:06 am

A. Let C(x) be the cost function.
the general form of C(x)= ax+b where a is the variable cost, x is the number of items and b is the fixed cost.
Here, 125=3a+b …..(1)
125=7a+b ….(2)
Suntracting (2) from (1) we get -4a=0 i.e. a=0 i.e. therefore b=125
i.e. C(x)=125

B. Revenure function: R(x)=25x where $25 is the selling price.

C. Profit Function: P(x)=R(x)-C(x)=25x-125

D. Break even Point:
R(x)=C(x) is the condition to find break event i.e. then there is no profit or loss.
25x=125 so, x=5

So if 5 items are sold i.e. when x=5 she will make neither profit nor loss which is break even point.

stephenwoodford
 
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