Thread: Differentiation - Could some one check my work

hi, im quite new to calculus and find it taxing but enjoyable. ive been given 4 questions to answer and have had a good go at them. how ever other class mates have got completely different answers to me , could some one pleae just check over them to make sure im going in the correct direct. many thanks
questions.doc

Hi

If find the same answers but you haven't finished the last exercice

Hi, thanks for the reply, as for the last question. What else do I have to do to it? Do I have to differentiate the previous differentiated formula and put my x values into that formula and use that answer as the x vale for the original formula? Many thanks.

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Originally Posted by djstar

Hi, thanks for the reply, as for the last question. What else do I have to do to it? Do I have to differentiate the previous differentiated formula and put my x values into that formula and use that answer as the x vale for the original formula? Many thanks.

You have found the volume V(x) = (2-2x).(2-2x).(x)
You must determine the value of the piece to be cut out to give the maximum volume
This can be done by diffrentiating V(x)

Thanks Ackbeet for the advice, ill remember that one.

ok so ive worked out the volume for the trough is

Volume = 4x3 – 8x2 +4x



[IMG]file:///C:/DOCUME%7E1/LIAMST%7E1/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image002.gif[/IMG]now i have found dy/dx = 12x2 – 8x +4


This has left me with a quadratic so i have put this into the quadratic equation



and have came up with two x answer's these are x=1 and x =0.33


do i put the 0.33 into the orignal equation 4(0.333 )– 8(0.332 )+4(0.33) to find the volume which i worked out as 0.592m2


Or do i have to differentiate the equation again





d2y/dx2 = 24x - 8


and substitue 0.33 into the euqation to give the answer -0.08


and then use this in the original formula 4(-0.083 )– 8(-0.082 )+4(-0.08) = -1.28 m2

V(x) = (2-2x).(2-2x).(x) = 4x(1-x)²
When differentiating V(x), you should not expand it
V'(x) = 4(1-x)²-8x(1-x) = 4(1-x)(1-3x)
It is easier to find the roots
V is defined for x between 0 and 1 therefore 1-x > 0
The sign of V'(x) is the same as 1-3x

thanks for the reply running-gag , im sorry if im a pain for all the replies, but im really new to calculus and im still at secondary schoollearning it so all of these notations and formulas i can find confusing.

could i just check with you that the correct answer for the volume of the trough is 0.592 metres cubed?

You are right
The maximum volume is obtained when x=1/3 and V(1/3) = 16/27 = 0.593 m^3