An open-topped cylindrical pot is to have volume 125 cm3. Determine the minimum possible amount of material used in making this pot? Neglect the thickness of the material as well as possible wastage. Give your answer accurate to 2 decimal places
In other words determine the minimum surface area for the given volume.
V = pi r^2 h
so h = 125/(pi r^2)
A = pi r^2 + 2 pi r h
A = pi r^2 + 2 pi r(125/(pi r^2))
A= pi r^2 + 250/r
dA/dr = 0 at min = 2 pi r -250/r^2
so
2 pi r = 250/r^2
pi r^3 =125
pi^(1/3) r = 5
r = 5/pi^(1/3)
To determine the minimum amount of material used in making the pot, we need to consider the surface area of the pot. Since the pot is open-topped and cylindrical, its volume can be given by the formula:
V = πr^2h
where V is the volume, r is the radius of the pot's base, and h is the height of the pot.
In this case, the volume is given as 125 cm^3, so we have:
125 = πr^2h
To find the minimum surface area, we need to minimize the amount of material used. The surface area of the cylinder is given by the formula:
A = 2πrh + πr^2
Since we want to minimize the amount of material, we can minimize the surface area. To do this, we can differentiate the surface area formula with respect to r and h and set them equal to zero to find the critical points.
Let's differentiate the equation with respect to r:
dA/dr = 2πh + 2πr
Setting this equal to zero:
2πh + 2πr = 0
2πr = -2πh
r = -h
Next, let's differentiate the equation with respect to h:
dA/dh = 2πr
Setting this equal to zero:
2πr = 0
r = 0
Now, let's consider the given conditions. Since the radius and height of the pot cannot be negative and the pot cannot have a radius of 0, we can disregard these critical points.
Therefore, the minimum surface area can be achieved when the pot has non-zero positive values for the radius and height.
Now we have two equations:
125 = πr^2h
A = 2πrh + πr^2
We need to find the minimum material used, which is equivalent to the minimum surface area. To minimize A, we can substitute the first equation into the second equation:
A = 2πrh + πr^2
= 2πr * (125 / πr^2) + πr^2
= 250/r + πr^2
To find the minimum, we can differentiate this equation with respect to r:
dA/dr = -250/r^2 + 2πr
Setting this equal to zero:
-250/r^2 + 2πr = 0
2πr = 250/r^2
2πr^3 = 250
r^3 = 125/π
r = (125/π)^(1/3)
Substituting this value back into the first equation to find the height:
125 = π(125/π)^(2/3) * h
h = 125 * π^(1/3) / (125/π)^(2/3)
Plugging these values back into the surface area equation to find the minimum surface area:
A = 2πr * h + πr^2
= 2π(125/π)^(1/3) * 125 * π^(1/3) / (125/π)^(2/3) + π * (125/π)^(2/3)
= 250 * (125/π)^(1/3) / (125/π)^(2/3) + (125/π)^(2/3)
= 250 * (125/π)^(1/3) * (125/π)^(1/3) / (125/π)^(2/3) + (125/π)^(2/3)
= 250 * (125/π)^(2/3) / (125/π)^(2/3) + (125/π)^(2/3)
= 250 + (125/π)^(2/3)
Calculating this value gives us:
A ≈ 250 + 10.52 ≈ 260.52 cm^2
Therefore, the minimum possible amount of material used in making the pot is approximately 260.52 cm^2.