The perimeter of the axial section of the cylinder is 6 dm. At what radius of the base of the cylinder

| August 9, 2021 | education

The perimeter of the axial section of the cylinder is 6 dm2. At what radius of the base of the cylinder will its lateral surface area be the largest?

Let us denote by R and H, respectively, the radius of the base and the height of the cylinder.
It is clear that the axial section of the cylinder has the shape of a rectangle with a base of 2 * R and a height of H. According to the specification, the perimeter of this rectangle is 6 dm. Using the formula for calculating the perimeter of a rectangle, we have 2 * (2 * R + H) = 6 dm, whence 2 * R + H = 3 dm.
As you know, the lateral surface Sside of the cylinder can be calculated by the formula Sside = 2 * π * R * H. The last equality in item 2 allows us to express H through R. We have H = 3 – 2 * R. Here and in what follows, for brevity, we omit unit of measurement (dm). Substitute H in its place in the formula for calculating the lateral surface. We have S side = 2 * π * R * (3 – 2 * R) = 6 * π * R – 4 * π * R ^ 2.
Consider the notation S side (R) = 6 * π * R – 4 * π * R ^ 2 as a function of R and examine this function for an extremum. Let us calculate the derivative of the function Sbok (R). We have (S side (R)) ꞌ = (6 * π * R – 4 * π * R ^ 2) ꞌ = 6 * π – 8 * π * R. Equating the derivative to zero 6 * π – 8 * π * R = 0, we find R = 3/4.
Since for R ∈ (0; 3/4) the value of the derivative 6 * π – 8 * π * R> 0, and for R ∈ (3/4; 3/2) the value of the derivative 6 * π – 8 * π * R < 0, then the function Sbok (R) takes its greatest value at R = 3/4.
Thus, with a radius of 3/4 dm = 0.75 dm = 7.5 cm, the area of ​​the lateral surface of the cylinder will be the largest.
Answer: With a radius equal to 0.75 dm = 7.5 cm, the area of ​​the lateral surface of the cylinder will be the largest.



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