IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: tmisterr on November 27, 2009, 01:22:49 pm
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if any one has the pure maths 1 text book for CIE, i need help with miscellaneous excersice 7 number 15, 16, and 17!
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Could you tell me the questions cause i bet i don't have the same book as you.
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Yes better ifyou post the question
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The costs of a firm which makes climbing boots are of two kind:
-Fixed cost (plant rates, office expenses): $2000 per week;
-Production costs (material, labour): $20 for each pair of boots made.
Market research suggest that if they price the boots at $30 a pair they will sell 500 pairs a week, but that at $55 a apair they will sell none at all; and between these values the graph of sales against profit is a straight line.
If they price boots at $x pair (30<x<55) (inclusive) find expressions for
a) weekly sales,
b) the weekly reciepts
c) the weekly costs
assuming that just enought boots are made
hence showing that the weeklly profit, $P, is given by
P=-20x2+ 1500x-2400.
Find the price at which the boots should be sold to maximise the profit.
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Sketch the graph of an even function f(x) which has a derivative at every point.
Let P be the point on the graph for which x=p (where p>0). Draw the tangent at P on your sketch. Also draw the tangetn at the point P' for which x=-p.
a) what is the relationship between the gradient at P' and the gradient at P? what can you deduce about the relationship between the gradient f'(p) and f'(-p)? what does this tell you about the derivative of an even function?
b) show the derivative of an odd function is even.
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When i get home
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omg
i had some very similiar kinda question.
it was about finding the maximum profit using derivatives.
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The costs of a firm which makes climbing boots are of two kind:
-Fixed cost (plant rates, office expenses): $2000 per week;
-Production costs (material, labour): $20 for each pair of boots made.
Market research suggest that if they price the boots at $30 a pair they will sell 500 pairs a week, but that at $55 a apair they will sell none at all; and between these values the graph of sales against profit is a straight line.
If they price boots at $x pair (30<x<55) (inclusive) find expressions for
a) weekly sales,
b) the weekly reciepts
c) the weekly costs
assuming that just enought boots are made
hence showing that the weeklly profit, $P, is given by
P=-20x2+ 1500x-2400.
Find the price at which the boots should be sold to maximise the profit.
a)sales=-20x+1100
b)receipts=sales*prce=x(-20x+1100)
c)costs=20x+2000
Surely something I have not understood. Are you sure " the graph of sales against profit is a straight line" is correct from the question?
f
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yes thats what the question says!!
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Stuck on 1 question for like 1.5 hours
i mean 1 hour 30 mins
lim t-->2 ( (t^3 + 3t^2 - 12t + 4) / (t^3 - 4t) )
[in english]
(t(cube) + 3t(square) - 12t + 4) divided by (t(cube) - 3t)
when the lim of t approaches 2
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L'hopitals rule. When i get home.
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i dont know that rule.
anyone knowing that over here.
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L'Hopitals rule: If f=g=0 at x=a them lim x-->a f/g=f'/g'
t^3 + 3t^2 - 12t + 4 =f so f'=3t^2+6t-12 and when t=2, f'=12
g=t^3-4t so g'=3t^2-4 and when t=2 g'=8
then lim t-->2 f/g=12/8=1.5
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but then where is it mentioned that f=g=0 ???
and can we use the derivative anytime we want to? (because its so easy :D)
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f=g has to be 0 else you cannot use the rule.
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aha
i love this one :D