Find the maximum volume of a cylindrical tin that has a total surface area of 600pi cm^2.
a = 2pi r^2 + 2pi rh = 600pi
h = (600pi - 2pi r^2)/(2pi*r)
= 300/r - r
v = pi r^2 h
= pi r^2 (300/r - r)
= 300pi r - pi r^3
dv/dr = 300pi - 3pi r^2
dv/dr = 0 when r = 10
so, h = 20
max v = pi * 100 * 20 = 200pi
To find the maximum volume of a cylindrical tin with a total surface area of 600π cm², we can use the properties of a cylinder.
The total surface area of a cylinder can be expressed as the sum of the areas of the two bases and the lateral surface area:
Total Surface Area = 2πr^2 + 2πrh
Where r is the radius of the base and h is the height of the cylinder.
In this case, the given total surface area is 600π cm². So, we have:
2πr^2 + 2πrh = 600π
We can simplify this equation by dividing both sides by 2π:
r^2 + rh = 300
Now, we need to express the volume of the cylinder in terms of r and h. The volume of a cylinder is given by:
Volume = πr^2h
To find the maximum volume, we need to maximize the function Volume = πr^2h subject to the constraint r^2 + rh = 300.
To simplify the problem, we can solve the constraint equation for h in terms of r:
rh = 300 - r^2
h = (300 - r^2)/r
Now, substitute this expression for h in the volume equation:
Volume = πr^2((300 - r^2)/r)
= πr(300 - r^2)
To find the maximum volume, we can differentiate the volume function with respect to r and set it equal to zero. Let's find the critical points:
d(Volume)/dr = π(300 - 3r^2) = 0
300 - 3r^2 = 0
3r^2 = 300
r^2 = 100
r = ±10
Since negative radii are not meaningful in this context, we consider r = 10.
Now, substitute this value of r into the constraint equation to find h:
10^2 + 10h = 300
100 + 10h = 300
10h = 200
h = 20
Therefore, the maximum volume of the cylindrical tin is achieved when the radius is 10 cm and the height is 20 cm.
To find the maximum volume of a cylindrical tin, we need to optimize the volume function subject to the constraint of the total surface area.
Let's denote the radius of the cylinder as "r" and the height of the cylinder as "h".
The formula for the total surface area of a cylinder is given by:
SA = 2πr^2 + 2πrh
Given that the total surface area is 600π cm^2, we can write the equation as:
600π = 2πr^2 + 2πrh
Now, we need to express the volume of the cylinder in terms of "r" and "h". The formula for the volume of a cylinder is:
V = πr^2h
To find the maximum volume, we need to differentiate the volume function with respect to either "r" or "h" and set the derivative equal to zero. This will give us the critical points, which we can then analyze to find the maximum.
Let's differentiate the volume equation with respect to "h":
dV/dh = πr^2
Now, we solve the constraint equation and differentiate the volume equation with respect to "r" using the same process.
From the constraint equation: 600π = 2πr^2 + 2πrh,
we can solve for "h" in terms of "r": h = (600 - 2r^2) / (2r)
Differentiating the volume equation (V = πr^2h) with respect to "r":
dV/dr = 2πrh + πr^2(dh/dr)
Substituting the expression for "h" from the constraint equation:
dV/dr = 2πr * [(600 - 2r^2) / (2r)] + πr^2(dh/dr)
dV/dr = πr(600 - 2r^2) / r + πr^2(dh/dr)
dV/dr = 300π - πr^2 + πr^2(dh/dr)
dV/dr = 300π
Since dV/dr is constant (300π), the derivative is not dependent on "r" or "h". This means that the volume function does not have a critical point or maximum value when differentiating with respect to "r". Therefore, to maximize the volume, we should use the constraint equation to solve for "h" in terms of "r" and substitute it into the volume equation.
Plug the expression for "h" back into the volume equation:
V = πr^2h
V = πr^2[(600 - 2r^2) / (2r)]
V = πr(600 - r^2) / 2
Now, to find the maximum volume, we need to find the critical points of this function. Let's differentiate the volume equation with respect to "r" and set it equal to zero:
dV/dr = π(600 - 3r^2) / 2 = 0
Solving for "r":
600 - 3r^2 = 0
3r^2 = 600
r^2 = 200
r = √(200)
r = 10√2
Substituting this value of "r" back into the expression for "h" in the constraint equation:
h = (600 - 2r^2) / (2r)
h = (600 - 2(10√2)^2) / (2(10√2))
h = (600 - 200) / (20√2)
h = 20 / √2
h = 10√2
So, the radius of the cylinder is 10√2 cm, and the height is also 10√2 cm.
Now, substitute these values into the volume equation to find the maximum volume:
V = πr(600 - r^2) / 2
V = π(10√2)(600 - (10√2)^2) / 2
V = 5000π cm^3
Therefore, the maximum volume of the cylindrical tin with a total surface area of 600π cm^2 is 5000π cm^3.