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notnek01

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samar444 said:
what about october november 2012 43 question 5 part b) i)?
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Width = Area / Length:

y = 5/x

Y = 6/(x+2)

y + Y = 1 --> 5/x + 6/(x+2) = 1

--> 5(x+2)/[x(x+2)] + 6x/[x(x+2)] = 1

--> [5(x+2)+6x]/[x(x+2)] = 1

--> 5(x+2) + 6x = x(x+2)

--> 5x + 10 + 6x = x^2 + 2x

--> x^2 - 9x - 10 = 0
 
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inquisitiveness

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Q.10d i and ii,we have to draw that,which I did,and I have attached the picture of what I drew which i checked more than once here,now for finding least cost of boxes,I know that method of substitu
ting the points in the unshaded area with the 5x + 2y but do we have to choose and try EACH and every point in the unshaded area?as the question is only for one mark,also,if you take point 6.5,6.5 and subtitute in the formula you would get 45.5 which is less than 47,does that mean we can never take any numbers other than whole numbers while subsituting in these kind of inequality questions?
thank you!!​
 
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inquisitiveness said:
Q.10d i and ii,we have to draw that,which I did,and I have attached the picture of what I drew which i checked more than once here,now for finding least cost of boxes,I know that method of substitu

ting the points in the unshaded area with the 5x + 2y but do we have to choose and try EACH and every point in the unshaded area?as the question is only for one mark,also,if you take point 6.5,6.5 and subtitute in the formula you would get 45.5 which is less than 47,does that mean we can never take any numbers other than whole numbers while subsituting in these kind of inequality questions?​
thank you!!​
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Large boxes cost more than small boxes so it's more important for x to be as small as possible. So focus on making x small first and then think about y.

Only the price for individual boxes is given so you can assume that the values of x and y must be whole numbers. This is usually the case for linear programming questions but you should always read the question to be sure.

So the smallest whole number of x in the region is 7 so that must be the solution along with y=6.

In an IGCSE exam, I recommend checking every point if you're not confident and have time.
 
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thank you i really appreciated!!,do you like work with IGCSE exam board or something lol we might as well start paying you :p
 
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notnek01 said:
Large boxes cost more than small boxes so it's more important for x to be as small as possible. So focus on making x small first and then think about y.

Only the price for individual boxes is given so you can assume that the values of x and y must be whole numbers. This is usually the case for linear programming questions but you should always read the question to be sure.

So the smallest whole number of x in the region is 7 so that must be the solution along with y=6.

In an IGCSE exam, I recommend checking every point if you're not confident and have time.
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btw do you try the points on the edge of unshaded area or all the points in the unshaded area including ones like in the centre?
 
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inquisitiveness said:
btw do you try the points on the edge of unshaded area or all the points in the unshaded area including ones like in the centre?
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This depends on whether the inequalities are strict (< or >) or not strict (<= or >=). A strict inequality should be drawn as a dashed line and any points on a dashed line are not included in the solution region.

Any points that lie on a solid line are included in the solution region.
 
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