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# C2- Applications of differentiation help! watch

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1. Please can you help me with the following questions. I missed a few lessons on this and this is just confusing me.
1) A man wishes to fence in a rectangular enclosure of are 128m^2. One side of the enlosure is formed by part of a brick wall already in position. What is the least possible length of fencing required for the other three sides?

2) The stamp counter at a post office is open from 9am to 3 pm daily. The average queuing time q, measured in minutes, at time x hours after opening is modelled by the equation q=1/4(-x^3+6x^2+20)
a) Show that using this model, the longest average queuing time occurs around 1 pm and find how long it is.
b) Find when the average queuing time is rising at its maximum rate, and show that this maximum rate of increase is 3 minutes/hour.

2. 1.)
Let x be the two sides and y the side parallel to the wall.
area=128=xy y=128/x
perimeter=P=2x+y
P=2x+128x^-1
dP/dx=2-128x^-2=0 (for a minimum)
2=128/x^2
2x^2=128
x^2=64
x=8 (must be positive as its a length)]
y=128/8=16
P=16+(2*8)=32
3. 2) The stamp counter at a post office is open from 9am to 3 pm daily. The average queuing time q, measured in minutes, at time x hours after opening is modelled by the equation q=1/4(-x^3+6x^2+20)
a) Show that using this model, the longest average queuing time occurs around 1 pm
Q = (6x2-x3+20)/4
dQ/dx = 3x - 3x2/4

Let dQ/dx = 0, so
=> 3x - 3x2/4 = 0
=> 3x = 3x2/4
=> 12 = 3x
:. x = 4
and find how long it is.
4 hours after opening, so 9 + 4 = 1pm.

sub (x = 4) into Q => (6(4)2 - (4)3+20)/4 = 13 min.
4. 2.)
a.)
q=1/4(-x^3+6x^2+20)
dq/dx=1/4(-3x^2+12x)=0 (for a maximum)
12x-3x^2=0
3x(4-x)=0
x=0 or 4
d2q/dx2=1/4(-6x+12)
If x=0 d2q/dx2 is +ve, so it is a minimum.
If x=4 d2q/dx2 is -ve, so it is a maximum.
9am+4 hours is 1pm.

b.)
d2q/dx2=1/4(-6x+12)=0 (for maximum)
12=6x x=2
Put this into dq/dx.
dq/dx=1/4(-3x^2+12x)
=1/4(-12+24)
=3 min/hour.
5. 2) The stamp counter at a post office is open from 9am to 3 pm daily. The average queuing time q, measured in minutes, at time x hours after opening is modelled by the equation q=1/4(-x^3+6x^2+20)
b) Find when the average queuing time is rising at its maximum rate, and show that this maximum rate of increase is 3 minutes/hour.
dQ/dx = 3x - 3x2/4
d2Q/dx2 = 3 - 3x/2

Let d2Q/dx2 = 0, so

=> 0 = 3 - 3x/2
=> 3 = 3x/2
=> x = 2

sub (x = 2) into dQ/dx => 3(2) - 3(2)2/4 = 3min.
6. thanks guys

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Updated: February 19, 2006
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