Dennis Marungu
Jun 11
A man has 75 pi centimeters squared sheet of metal and which is to make an open cylindrical container with it. Find The Dimension of the tin for which the volume will be maximum?
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Jun 11
Let the radius of the cylindrical container be r and the height be h.
The surface area of the cylindrical container is given by:
2πrh + πr^2
Given that the total surface area of the metal sheet is 75π cm^2, we have:
2πrh + πr^2 = 75π
Simplifying, we get:
2rh + r^2 = 75
The volume of the cylindrical container is given by:
V = πr^2h
We want to maximize the volume, so we need to express the volume in terms of one variable. From the surface area equation, we can express h in terms of r:
h = (75 - r^2) / (2r)
Substitute this expression for h into the volume equation:
V = πr^2 * (75 - r^2) / (2r)
V = π/2 * r * (75r - r^3)
To find the maximum volume, we need to find the critical points of V. Taking the derivative of V with respect to r and setting it equal to 0:
dV/dr = π/2 * (75 - 3r^2) = 0
75 - 3r^2 = 0
r^2 = 25
r = 5
Therefore, the radius of the cylindrical container for which the volume will be maximum is 5 cm. To find the corresponding height, substitute r = 5 into the expression for h:
h = (75 - 5^2) / (2*5)
h = 5
So, the dimensions of the tin for which the volume will be maximum are:
Radius = 5 cm
Height = 5 cm
The surface area of the cylindrical container is given by:
2πrh + πr^2
Given that the total surface area of the metal sheet is 75π cm^2, we have:
2πrh + πr^2 = 75π
Simplifying, we get:
2rh + r^2 = 75
The volume of the cylindrical container is given by:
V = πr^2h
We want to maximize the volume, so we need to express the volume in terms of one variable. From the surface area equation, we can express h in terms of r:
h = (75 - r^2) / (2r)
Substitute this expression for h into the volume equation:
V = πr^2 * (75 - r^2) / (2r)
V = π/2 * r * (75r - r^3)
To find the maximum volume, we need to find the critical points of V. Taking the derivative of V with respect to r and setting it equal to 0:
dV/dr = π/2 * (75 - 3r^2) = 0
75 - 3r^2 = 0
r^2 = 25
r = 5
Therefore, the radius of the cylindrical container for which the volume will be maximum is 5 cm. To find the corresponding height, substitute r = 5 into the expression for h:
h = (75 - 5^2) / (2*5)
h = 5
So, the dimensions of the tin for which the volume will be maximum are:
Radius = 5 cm
Height = 5 cm
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Google Gemini AI
Jun 11
Given:
Sheet of metal = 75π cm²
To find:
Dimensions of container for maximum volume
Solution:
Let the radius of base be "r" cm and height be "h" cm.
Surface area of sheet = Curved surface area of cylinder + 2(Area of base)
75π = 2πrh + 2πr²
Rearranging the equation:
h = (75π - 2πr²)/2πr
Volume of cylinder:
V = πr²h
Substituting the value of h:
V = πr²((75π - 2πr²)/2πr)
Simplifying:
V = 37.5π - π²r³
To find the maximum volume, we need to find the critical points:
dV/dr = 3πr² - 3π³r
Equating the derivative to zero:
3πr²(1 - r) = 0
Hence, r = 0 or r = 1. But r cannot be zero (no base).
Therefore, r = 1 cm
Substituting the value of r back into the equation for h:
h = (75π - 2π(1)²) / 2π(1)
h = 36.5 cm
Therefore, the dimensions of the tin for maximum volume are:
Radius of base (r) = 1 cm
Height (h) = 36.5 cm
Sheet of metal = 75π cm²
To find:
Dimensions of container for maximum volume
Solution:
Let the radius of base be "r" cm and height be "h" cm.
Surface area of sheet = Curved surface area of cylinder + 2(Area of base)
75π = 2πrh + 2πr²
Rearranging the equation:
h = (75π - 2πr²)/2πr
Volume of cylinder:
V = πr²h
Substituting the value of h:
V = πr²((75π - 2πr²)/2πr)
Simplifying:
V = 37.5π - π²r³
To find the maximum volume, we need to find the critical points:
dV/dr = 3πr² - 3π³r
Equating the derivative to zero:
3πr²(1 - r) = 0
Hence, r = 0 or r = 1. But r cannot be zero (no base).
Therefore, r = 1 cm
Substituting the value of r back into the equation for h:
h = (75π - 2π(1)²) / 2π(1)
h = 36.5 cm
Therefore, the dimensions of the tin for maximum volume are:
Radius of base (r) = 1 cm
Height (h) = 36.5 cm
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