The height and the radius that minimize the material needed to manufacture the can is 4.18336 cm and 4.18337 cm respectively.
Computed using differentiation.
We assume the radius of the can be r cm, and its height to be h cm.
We are given that the can is to hold 230 cm³ of liquid, thus the volume of the can is:
V = 230,
or, πr²h = 230 {Since, the volume of a cylinder is πr²h, where r is the radius, and h is the height},
or, h = 230/(πr²) ... (i) .
We are asked to find the height and radius that minimize the amount of the material.
To calculate the amount of the material, we calculate the surface area of the can (A).
The surface area of the can = area of the base + area of the side,
or, A = πr² + 2πrh = πr(r + 2h).
Substituting h = 230/(πr²), we get:
A = πr(r + 2{230/(πr²)}),
or, A = {πr(πr³ + 460)}/πr² = πr² + 460/r.
Differentiating both sides with respect to the radius r, we get:
dA/dr = 2πr - 460/r² ...(ii)
To find the point of inflection, we equate this to zero, to get:
2πr - 460/r² = 0,
or, 2πr = 460/r²,
or, r³ = 460/2π = 73.2113,
or, r = 4.18337 cm.
To check whether the point of inflection shows the point of maximum or minimum, we differentiate (ii) with respect to the radius r, to get:
d²A/dr² = 2π + 920/r³, which is always positive when the radius r is positive.
Thus, the area is minimum when the radius r = 4.18337 cm.
Height, h = 230/(πr²) = 4.18336 cm.
Thus, the height and the radius that minimize the material needed to manufacture the can is 4.18336 cm and 4.18337 cm respectively.
Computed using differentiation.
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