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C2 Questions on Differentiation! I need help!!!
I've got a test soon, and have been trying to revise differentiation, but I need help with these application of diff. Qs! Could you guys pleaseee help me out?!?!
9C
1) A rectangular garden is fenced on three sides, and the house forms the fourth side of the rectangle.
Given that the total length of the fence is 80 m show that the area, A, of the garden is given by the formula A=y(80-2y), where y is the distance from the house to the end of the garden.
Given that the are is a maximum for this length of fence, find the dimensions of the enclosed garden, and the area which is enclosed.
9C
1) A rectangular garden is fenced on three sides, and the house forms the fourth side of the rectangle.
Given that the total length of the fence is 80 m show that the area, A, of the garden is given by the formula A=y(80-2y), where y is the distance from the house to the end of the garden.
Given that the are is a maximum for this length of fence, find the dimensions of the enclosed garden, and the area which is enclosed.
Reply 1
*girlie*
I've got a test soon, and have been trying to revise differentiation, but I need help with these application of diff. Qs! Could you guys pleaseee help me out?!?!
9C
1) A rectangular garden is fenced on three sides, and the house forms the fourth side of the rectangle.
Given that the total length of the fence is 80 m show that the area, A, of the garden is given by the formula A=y(80-2y), where y is the distance from the house to the end of the garden.
Given that the are is a maximum for this length of fence, find the dimensions of the enclosed garden, and the area which is enclosed.
9C
1) A rectangular garden is fenced on three sides, and the house forms the fourth side of the rectangle.
Given that the total length of the fence is 80 m show that the area, A, of the garden is given by the formula A=y(80-2y), where y is the distance from the house to the end of the garden.
Given that the are is a maximum for this length of fence, find the dimensions of the enclosed garden, and the area which is enclosed.
Hehe, i've done this question about 3 times now and i'm sick of it. nevertheless i'll help you.
Okay, let one side of the rectangle (the distance between the house and the end of the garden) be ym and the other side be xm.
Perimeter of fence
2y + x = 80m
x = 80 - 2y
Area of the garden = A = xy = y(80 - 2y)
Area is a maximum
=> dA/dy = 0
=> 80 - 4y = 0
=> y = 80/4 = 20m
x = 80 - 2y
x = 80 - 2(20) = 40
Area of garden = xy = 40*20 = 800m^2
Reply 2
*girlie*
1) A rectangular garden is fenced on three sides, and the house forms the fourth side of the rectangle.
Given that the total length of the fence is 80 m show that the area, A, of the garden is given by the formula A=y(80-2y), where y is the distance from the house to the end of the garden.
Given that the area is a maximum for this length of fence, find the dimensions of the enclosed garden, and the area which is enclosed.
Given that the total length of the fence is 80 m show that the area, A, of the garden is given by the formula A=y(80-2y), where y is the distance from the house to the end of the garden.
Given that the area is a maximum for this length of fence, find the dimensions of the enclosed garden, and the area which is enclosed.
Area = Length * Breadth
Length = y
2y + Breadth = 80 (Only 3 sides to consider remember)
-> Breadth = 80 - 2y
-> Area = Length * Breadth = y(80 - 2y)
Area = 80y - 2y^2
-> dA/dy = 80 - 4y
At max. point i.e.) When Max. Area occurs:
-> 80 - 4y = 0
-> 4y = 80
-> y = 20
Therefore:
Length = 20
Breadth = 80 - 2(20) = 80 - 40 = 40
-> Max. Area = 20 * 40 = 800 Sq. units
Right, part one:
Let the width of the garden (the length of fence that is not y) be X.
Therefore: 2y + X = 80
so: X = 80 - 2y
Area of rectangle = length x width = yX = y(80 - 2y)
Part Two:
A = 80y - 2(y^2)
dA/dy = 80 - 4y
A is maximum, so dA/dy = 0
So: 80 - 4y = 0
so: 80 = 4y
so: y = 20
As defined earlier: X = 80 - 2y = 80 - 2(20) = 40
so: A = 40 x 20 = 800 m^2
I bet someone has beaten me to this.
Let the width of the garden (the length of fence that is not y) be X.
Therefore: 2y + X = 80
so: X = 80 - 2y
Area of rectangle = length x width = yX = y(80 - 2y)
Part Two:
A = 80y - 2(y^2)
dA/dy = 80 - 4y
A is maximum, so dA/dy = 0
So: 80 - 4y = 0
so: 80 = 4y
so: y = 20
As defined earlier: X = 80 - 2y = 80 - 2(20) = 40
so: A = 40 x 20 = 800 m^2
I bet someone has beaten me to this.
If you have a function y = f(x), then this function tells you how y changes when you vary x. Now, the operation of differentiation tells you more: the rate at which y changes when you vary x.
To get the expression for the area...
Let y be the length of the garden. The sum of the two sides of the garden is therefore 2y, and so the length of fence at the far end of the garden must be 80-2y. Area = length x width so A = y(80-2y).
So this function tells you how the area of the garden changes as you vary the length of the garden. By inspecting this equation, you can see that if you start with y=0 and increase it by a metre each time, A will go through the following values:
y*(80-y) = A
0*80 = 0
1*78 = 78
2*76 = 152
5*70 = 350
10*60 = 600
20*40 = 800
30*20 = 600
35*10 = 350
40*0 = 0
(if y>40, this produces negative areas which you can ignore)
So you can see that the area first increases and then decreases and y is varied. The maximum value of the area will be when A switches from increasing to decreasing. This is where differentiation comes in.
If the derivative dA/dy is positive, then A is increasing. If dA/dy is negative then A is decreasing. So the point at which it switches from increasing to decreasing is where dA/dy = 0.
So, find dA/dy.
A = 80y - 2y^2 (multiply each term by the power of y then lower the power by one)
dA/dy = 80 - 4y
Set this equal to zero to find the point at which the function changes from increasing to decreasing...
dA/dy = 0
80 - 4y = 0
4y = 80
y = 20
So the value of y at which the area stops increasing and starts decreasing (when the derivative of A with respect to y is zero), is 20.
So we know the maximum area occurs when y = 20.
Maximum area = y(80-2y) when y = 20
A(max) = 20*(80-2*20)
A(max) = 20*(40)
A(max) = 800m^2
Which is the highest value we happened across when doing the trials earlier.
To get the expression for the area...
Let y be the length of the garden. The sum of the two sides of the garden is therefore 2y, and so the length of fence at the far end of the garden must be 80-2y. Area = length x width so A = y(80-2y).
So this function tells you how the area of the garden changes as you vary the length of the garden. By inspecting this equation, you can see that if you start with y=0 and increase it by a metre each time, A will go through the following values:
y*(80-y) = A
0*80 = 0
1*78 = 78
2*76 = 152
5*70 = 350
10*60 = 600
20*40 = 800
30*20 = 600
35*10 = 350
40*0 = 0
(if y>40, this produces negative areas which you can ignore)
So you can see that the area first increases and then decreases and y is varied. The maximum value of the area will be when A switches from increasing to decreasing. This is where differentiation comes in.
If the derivative dA/dy is positive, then A is increasing. If dA/dy is negative then A is decreasing. So the point at which it switches from increasing to decreasing is where dA/dy = 0.
So, find dA/dy.
A = 80y - 2y^2 (multiply each term by the power of y then lower the power by one)
dA/dy = 80 - 4y
Set this equal to zero to find the point at which the function changes from increasing to decreasing...
dA/dy = 0
80 - 4y = 0
4y = 80
y = 20
So the value of y at which the area stops increasing and starts decreasing (when the derivative of A with respect to y is zero), is 20.
So we know the maximum area occurs when y = 20.
Maximum area = y(80-2y) when y = 20
A(max) = 20*(80-2*20)
A(max) = 20*(40)
A(max) = 800m^2
Which is the highest value we happened across when doing the trials earlier.
Oh my, so many people helped out! Thanks a lot!!!
I have more...:-(
2) A closed cylinder has total surface area equal to 600 Pi (can’t find symbol for it). Show that the volume, V cm^3, of this cylinder is given by the formula V=300 Pi r – Pi r^3, where r cm is the radius of the cylinder. Find the maximum volume of such a cylinder.
I have more...:-(
2) A closed cylinder has total surface area equal to 600 Pi (can’t find symbol for it). Show that the volume, V cm^3, of this cylinder is given by the formula V=300 Pi r – Pi r^3, where r cm is the radius of the cylinder. Find the maximum volume of such a cylinder.
*girlie*
Oh my, so many people helped out! Thanks a lot!!!
I have more...:-(
2) A closed cylinder has total surface area equal to 600 Pi (can’t find symbol for it). Show that the volume, V cm^3, of this cylinder is given by the formula V=300 Pi r – Pi r^3, where r cm is the radius of the cylinder. Find the maximum volume of such a cylinder.
I have more...:-(
2) A closed cylinder has total surface area equal to 600 Pi (can’t find symbol for it). Show that the volume, V cm^3, of this cylinder is given by the formula V=300 Pi r – Pi r^3, where r cm is the radius of the cylinder. Find the maximum volume of such a cylinder.
The surface area is given by the formula
2 pi r^2 + 2 pi r h = 600 pi
From this you can find h in terms of r.
Then V = pi r^2 h
which you can now write as a function of r alone.
Find when dV/dr = 0 to get the maximum
9D
7) A cylindrical biscuit tin has a close-fitting lid which overlaps the tin by 1 cm. The radii of the tin and the lid are both x cm. The tin and the lid are made from a thin sheet of metal of area 80 Pi cm^2 and there is no wastage. The volume of the tin is V cm^3.
a)Show that V=Pi(40x-x^2-x^3).
Given that x can vary:
b)Use differentiation to find the positive value of x for which V is stationary.
c)Prove this value of x gives a maximum value of V.
d)Find this maximum value of V.
e)Determine the percentage of the sheet metal used in the lid when V is a maximum.
I really appreaciate the help!!! (and i need it!)
7) A cylindrical biscuit tin has a close-fitting lid which overlaps the tin by 1 cm. The radii of the tin and the lid are both x cm. The tin and the lid are made from a thin sheet of metal of area 80 Pi cm^2 and there is no wastage. The volume of the tin is V cm^3.
a)Show that V=Pi(40x-x^2-x^3).
Given that x can vary:
b)Use differentiation to find the positive value of x for which V is stationary.
c)Prove this value of x gives a maximum value of V.
d)Find this maximum value of V.
e)Determine the percentage of the sheet metal used in the lid when V is a maximum.
I really appreaciate the help!!! (and i need it!)
Reply 9
*girlie*
9D
7) A cylindrical biscuit tin has a close-fitting lid which overlaps the tin by 1 cm. The radii of the tin and the lid are both x cm. The tin and the lid are made from a thin sheet of metal of area 80 Pi cm^2 and there is no wastage. The volume of the tin is V cm^3.
a)Show that V=Pi(40x-x^2-x^3).
Given that x can vary:
b)Use differentiation to find the positive value of x for which V is stationary.
c)Prove this value of x gives a maximum value of V.
d)Find this maximum value of V.
e)Determine the percentage of the sheet metal used in the lid when V is a maximum.
I really appreaciate the help!!! (and i need it!)
7) A cylindrical biscuit tin has a close-fitting lid which overlaps the tin by 1 cm. The radii of the tin and the lid are both x cm. The tin and the lid are made from a thin sheet of metal of area 80 Pi cm^2 and there is no wastage. The volume of the tin is V cm^3.
a)Show that V=Pi(40x-x^2-x^3).
Given that x can vary:
b)Use differentiation to find the positive value of x for which V is stationary.
c)Prove this value of x gives a maximum value of V.
d)Find this maximum value of V.
e)Determine the percentage of the sheet metal used in the lid when V is a maximum.
I really appreaciate the help!!! (and i need it!)
7)a Area of lid and tin
= pi*x^2 + 2pi*x*1 + pi*x^2 + 2pi*y = 80picm^2 (where y is the height of the tin)
Right, okay before I continue, I'll explain each part.
pi*x^2 = area of lid
2pi*x*1 = area of 1cm overlap
pi*x^2 = area of base of the tin
2pi*y = the area of the height of the tin
80pi = total area which is given in the question
=> 80pi - 2pi*x^2 - 2pi*x = 2pi*x*y
=> y = (80pi - 2pi*x^2 - 2pi*x)/(2pi*x)
=> y = 40/x - x - 1
Volume of tin (V)
= pi*x^2*y
= pi*x^2(40/x - x - 1)
= 40pi*x - pi*x^3 - pi*x^2
= pi(40x - x^2 - x^3)
Reply 10
7)b
When x is stationary, dV/dx = 0
=> 40pi - 3pi*x^2 - 2pi*x = 0
=> pi(40 - 3x^2 - 2x) = 0
=> 40 - 3x^2 - 2x = 0
=> (10 - 3x)(4 + x) = 0
=> x = 10/3 or x = -4
x > 0 => x = 10/3
When x is stationary, dV/dx = 0
=> 40pi - 3pi*x^2 - 2pi*x = 0
=> pi(40 - 3x^2 - 2x) = 0
=> 40 - 3x^2 - 2x = 0
=> (10 - 3x)(4 + x) = 0
=> x = 10/3 or x = -4
x > 0 => x = 10/3
Reply 12
7)c
If V is a maximum, d^2A/dx^2 < 0
=> -6x - 2pi < 0
x = 10/3
-6(10/3) - 2pi < 0
Therefore V is a maximum when x = 10/3
7)d
V = pi(40x - x^2 - x^3)
x = 10/3
=> V = pi(400/3 - 100/9 - 1000/27)
=> V = 400pi/3 - 100pi/9 - 1000pi/27
=> V = 3600pi/27 - 300pi/27 - 1000pi/27
=> V = (3600pi - 300pi - 1000pi)/(27)
=> V = (2300pi/27)cm^2
7)e
Area of lid when x is a maximum, ie when x = 10/3
= pi*x^2 + 2pi*x
= pi*(10/3)^2 + 2pi(10/3)
= 100pi/9 + 20pi/3
= 100pi/9 + 60pi/9 = (160pi/9)cm^2
Percentage of sheet metal used
= (160pi/9)/(80pi)*100
= (160pi/720pi)*100
= (2/9)*100
= (200/9)%
= (22 and 2/9)%
If V is a maximum, d^2A/dx^2 < 0
=> -6x - 2pi < 0
x = 10/3
-6(10/3) - 2pi < 0
Therefore V is a maximum when x = 10/3
7)d
V = pi(40x - x^2 - x^3)
x = 10/3
=> V = pi(400/3 - 100/9 - 1000/27)
=> V = 400pi/3 - 100pi/9 - 1000pi/27
=> V = 3600pi/27 - 300pi/27 - 1000pi/27
=> V = (3600pi - 300pi - 1000pi)/(27)
=> V = (2300pi/27)cm^2
7)e
Area of lid when x is a maximum, ie when x = 10/3
= pi*x^2 + 2pi*x
= pi*(10/3)^2 + 2pi(10/3)
= 100pi/9 + 20pi/3
= 100pi/9 + 60pi/9 = (160pi/9)cm^2
Percentage of sheet metal used
= (160pi/9)/(80pi)*100
= (160pi/720pi)*100
= (2/9)*100
= (200/9)%
= (22 and 2/9)%
Reply 13
*girlie*
Thank you!!! Could you pls help me with Q4, which i have already posted? I dont understand, do you just do these Qs just for fun??
i'm doing them mainly for revision.
Reply 14
4) A shape consists of a rectangular base with a semicircular top. Given that the perimeter of the shape is 40 cm, show that its area, A cm^2, is given by the formula A=40r – 2r^2 – Pi r^2 / 2, where r cm is the radius of the semicircle. Find the maximum value for this area.
Let the width of the rectangle be x cm
The length of the rectangle is 2r cm
Perimeter = (2r + 2x + pi*r)cm = 40cm
=> 2x = 40 - pi*r - 2r
=> x = 20 - 0.5pi*r - r
Area of shape
= 2rx + 0.5pi*r^2
= 2r(20 - 0.5pi*r - r) + 0.5pi*r^2
= 40r - pi^2 - 2r^2 + 0.5pi*r^2
= 40r - 2r^2 - 0.5pi*r^2
Maximum value for area
dA/dr = 0
=> 40 - 4r - pi*r = 0
=> pi*r + 4r = 40
=> r(pi + 4) = 40
=> r = 40/(pi + 4)
The algebra gets a little bit complicated here.
A = 40r - 2r^2 - (pi*r^2)/2
A = 40[40/(pi+4)] - 2[40/(pi+4)]^2 - 0.5pi[40/(pi+4)]^2
A = 40[40/(pi+4)] - 2[40/(pi+4)][40/(pi+4)] - 0.5pi[40/(pi+4)][40/(pi+4)]
A = [40/(pi+4)][40 - 80/(pi+4) - 20pi/(pi+4)]
A = [40/(pi+4)][40(pi+4)/(pi+4) - 80/(pi+4) - 20pi/(pi+4)]
A = [40/(pi+4)][(40pi + 160 - 80 - 20pi)/(pi+4)]
A = [40/(pi+4)][(20pi + 80)/(pi+4)]
A = [40/(pi+4)][20(pi+4)/(pi+4)]
A = [40/(pi+4)][20]
A = 800/(pi+4) cm^2
Let the width of the rectangle be x cm
The length of the rectangle is 2r cm
Perimeter = (2r + 2x + pi*r)cm = 40cm
=> 2x = 40 - pi*r - 2r
=> x = 20 - 0.5pi*r - r
Area of shape
= 2rx + 0.5pi*r^2
= 2r(20 - 0.5pi*r - r) + 0.5pi*r^2
= 40r - pi^2 - 2r^2 + 0.5pi*r^2
= 40r - 2r^2 - 0.5pi*r^2
Maximum value for area
dA/dr = 0
=> 40 - 4r - pi*r = 0
=> pi*r + 4r = 40
=> r(pi + 4) = 40
=> r = 40/(pi + 4)
The algebra gets a little bit complicated here.
A = 40r - 2r^2 - (pi*r^2)/2
A = 40[40/(pi+4)] - 2[40/(pi+4)]^2 - 0.5pi[40/(pi+4)]^2
A = 40[40/(pi+4)] - 2[40/(pi+4)][40/(pi+4)] - 0.5pi[40/(pi+4)][40/(pi+4)]
A = [40/(pi+4)][40 - 80/(pi+4) - 20pi/(pi+4)]
A = [40/(pi+4)][40(pi+4)/(pi+4) - 80/(pi+4) - 20pi/(pi+4)]
A = [40/(pi+4)][(40pi + 160 - 80 - 20pi)/(pi+4)]
A = [40/(pi+4)][(20pi + 80)/(pi+4)]
A = [40/(pi+4)][20(pi+4)/(pi+4)]
A = [40/(pi+4)][20]
A = 800/(pi+4) cm^2
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