markdumbleton
New member
- Joined
- Mar 13, 2011
- Messages
- 1
hi, im quite new to calculus and find it taxing but enjoyable. ive been given a question to answer and have had a good go at it. 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
1. A through is 2m square, find a formula for the volume and determine the value of the square corner piece to be cut out to give the maximum volume when the sheet of metal is folded into a trough.
Volume = (2-2x).(2-2x).(x)
Volume = 4x^3 – 8x^2 +4x
i have found dy/dx = 12x^2 – 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 original equation 4(0.333 )^3– 8(0.332 )^2+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 substitute 0.33 into the equation to give the answer -0.08
and then use this in the original formula 4(-0.083 )^3– 8(-0.082 )^2+4(-0.08) = -1.28 m2
any help would be greatly appreciated.
1. A through is 2m square, find a formula for the volume and determine the value of the square corner piece to be cut out to give the maximum volume when the sheet of metal is folded into a trough.
Volume = (2-2x).(2-2x).(x)
Volume = 4x^3 – 8x^2 +4x
i have found dy/dx = 12x^2 – 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 original equation 4(0.333 )^3– 8(0.332 )^2+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 substitute 0.33 into the equation to give the answer -0.08
and then use this in the original formula 4(-0.083 )^3– 8(-0.082 )^2+4(-0.08) = -1.28 m2
any help would be greatly appreciated.