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C2 Application of Differentiation Qs!!! Pls help me!
9) A rope of length 60 m is used to fence off a rectangular area, one side of which needs rope because there is a wall. If x is the width of the rectangle, then:
a)Find an expression for the length of the rectangle (in terms of x).
b)Find an expression for the area, A, of the rectangle, in terms of x.
c)Hence find the value of x which would maximise the area and find the area that this would create.
d)Show that it is a maximum.
12) An open top cylinder has a total external surface area of 192 Pi cm^2. If r is the base radius and h is the height, then:
a)Find an expression for h in terms of r.
b)Hence how that the volume, V, can be given as: V= Pi/2 (192-r^3).
c)Use this to find the maximum possible volume of the cylinder, showing that this is indeed the maximum value.
a)Find an expression for the length of the rectangle (in terms of x).
b)Find an expression for the area, A, of the rectangle, in terms of x.
c)Hence find the value of x which would maximise the area and find the area that this would create.
d)Show that it is a maximum.
12) An open top cylinder has a total external surface area of 192 Pi cm^2. If r is the base radius and h is the height, then:
a)Find an expression for h in terms of r.
b)Hence how that the volume, V, can be given as: V= Pi/2 (192-r^3).
c)Use this to find the maximum possible volume of the cylinder, showing that this is indeed the maximum value.
Reply 1
(9)
(a)
60 - 2x
(b)
A
= width*length
= x(60 - 2x)
= 60x - 2x^2
(c)
dA/dx = 60 - 4x
The only stationary point of A occurs at x = 15.
(d)
d^2A/dx^2 = -4 < 0
So x = 15 maximises A.
--
(12)
(a)
192pi
= total surface area
= base surface area + curved surface area
= pi r^2 + 2pi r h
h
= (192 - r^2)/(2r)
= (1/2)(192/r - r)
(b)
V
= pi r^2 h
= (1/2)pi r^2(192/r - r)
= (1/2)pi (192r - r^3)
(c)
dV/dr = (1/2)pi (192 - 3r^2)
The only stationary point of V occurs at r = sqrt(192/3) = 8.
At r = 8, d^2V/dr^2 = -3pi*r is negative. So r = 8 maximises V.
(a)
60 - 2x
(b)
A
= width*length
= x(60 - 2x)
= 60x - 2x^2
(c)
dA/dx = 60 - 4x
The only stationary point of A occurs at x = 15.
(d)
d^2A/dx^2 = -4 < 0
So x = 15 maximises A.
--
(12)
(a)
192pi
= total surface area
= base surface area + curved surface area
= pi r^2 + 2pi r h
h
= (192 - r^2)/(2r)
= (1/2)(192/r - r)
(b)
V
= pi r^2 h
= (1/2)pi r^2(192/r - r)
= (1/2)pi (192r - r^3)
(c)
dV/dr = (1/2)pi (192 - 3r^2)
The only stationary point of V occurs at r = sqrt(192/3) = 8.
At r = 8, d^2V/dr^2 = -3pi*r is negative. So r = 8 maximises V.
Reply 2
*girlie*
OP
5
Thank you so much!! What about these:
13) A square card of side 6 cm has four squares of side x cm cut from each corner. The remaining card is folded to make an open box of height x cm.
a)Show that volume V of the open box is given by V=x(6-2x)^2 (I just don’t understand why it’s not x(6-2x)??
b)Find the maximum possible volume of the box. (The answer is 16cm^3, but I keep getting 396cm^2?!
21*) A rectangular sheet of tin cardboard is 80 cm by 50 cm. A square of side x centimetres is cut away from each corner of the sheet which is then folded to form an open rectangular box of volume y cubic centimetres. Show that:
y=4000x – 260x^2 + 4x^3.
Given that x varies, find the greatest volume of the box.
17) The slant height of a cone is 18 cm.
a)Find the volume of the cone in terms of h.
b)Hence find the exact value of the minimum volume of the cone.
13) A square card of side 6 cm has four squares of side x cm cut from each corner. The remaining card is folded to make an open box of height x cm.
a)Show that volume V of the open box is given by V=x(6-2x)^2 (I just don’t understand why it’s not x(6-2x)??
b)Find the maximum possible volume of the box. (The answer is 16cm^3, but I keep getting 396cm^2?!
21*) A rectangular sheet of tin cardboard is 80 cm by 50 cm. A square of side x centimetres is cut away from each corner of the sheet which is then folded to form an open rectangular box of volume y cubic centimetres. Show that:
y=4000x – 260x^2 + 4x^3.
Given that x varies, find the greatest volume of the box.
17) The slant height of a cone is 18 cm.
a)Find the volume of the cone in terms of h.
b)Hence find the exact value of the minimum volume of the cone.
Reply 3
10
13.
a) V = width * length * height = (6-2x) * (6-2x) * x = x(6-2x)²
b)
dV/dx = (6-2x)² - 4x(6-2x) = 0
6-2x = 4x
x = 1
=> V = 1(6-2)² = 16 cm³
21.
y = width * length * height = (50-2x) * (80-2x) * x = 4000x - 260x² + 4x³
dy/dx = 4000 - 520x + 12x² = 0
=> (3x-100)(x-10) = 0
=> x=10 or x=100/3
So, take x = 10, then: y = 18000.
17.
a)
Pythagoras:
18² = r² + h², where r is the radius and h is the height.
V = pi.r².h= pi(18²-h²)h = 324pi.h - h³pi
b)
dV/dh = 324pi - 3h²pi = 0
=> h = sqrt(108)
V = 324sqrt(108) pi - [sqrt(108)]³ pi
Hopefully no mistakes crept in.
a) V = width * length * height = (6-2x) * (6-2x) * x = x(6-2x)²
b)
dV/dx = (6-2x)² - 4x(6-2x) = 0
6-2x = 4x
x = 1
=> V = 1(6-2)² = 16 cm³
21.
y = width * length * height = (50-2x) * (80-2x) * x = 4000x - 260x² + 4x³
dy/dx = 4000 - 520x + 12x² = 0
=> (3x-100)(x-10) = 0
=> x=10 or x=100/3
So, take x = 10, then: y = 18000.
17.
a)
Pythagoras:
18² = r² + h², where r is the radius and h is the height.
V = pi.r².h= pi(18²-h²)h = 324pi.h - h³pi
b)
dV/dh = 324pi - 3h²pi = 0
=> h = sqrt(108)
V = 324sqrt(108) pi - [sqrt(108)]³ pi
Hopefully no mistakes crept in.
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