A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. The total volume of the solid is 10 cubic centimeters. Find the radius of the cylinder that produces the minimum surface area. (Round your answer to two decimal places.)

Answer:

[tex]r = 1.34[/tex]

Step-by-step explanation:

Given

Solid = Cylinder + 2 hemisphere

[tex]Volume = 10cm^3[/tex]

Required

Determine the radius (r) that minimizes the surface area

First, we need to determine the volume of the shape.

Volume of Cylinder (V1) is:

[tex]V_1 = \pi r^2h[/tex]

Volume of 2 hemispheres (V2) is:

[tex]V_2 = \frac{2}{3}\pi r^3 +\frac{2}{3}\pi r^3[/tex]

[tex]V_2 = \frac{4}{3}\pi r^3[/tex]

Volume of the solid is:

[tex]V = V_1 + V_2[/tex]

[tex]V = \pi r^2h + \frac{4}{3}\pi r^3[/tex]

Substitute 10 for V

[tex]10 = \pi r^2h + \frac{4}{3}\pi r^3[/tex]

Next, we make h the subject

[tex]\pi r^2h = 10 - \frac{4}{3}\pi r^3[/tex]

Solve for h

[tex]h = \frac{10}{\pi r^2} - \frac{\frac{4}{3}\pi r^3 }{\pi r^2}[/tex]

[tex]h = \frac{10}{\pi r^2} - \frac{4\pi r^3 }{3\pi r^2}[/tex]

[tex]h = \frac{10}{\pi r^2} - \frac{4r }{3}[/tex]

Next, we determine the surface area

Surface area (A1) of the cylinder:

Note that the cylinder is covered by the 2 hemisphere.

So, we only calculate the surface area of the curved surface.

i.e.

[tex]A_1 = 2\pi rh[/tex]

Surface Area (A2) of 2 hemispheres is:

[tex]A_2 = 2\pi r^2+2\pi r^2[/tex]

[tex]A_2 = 4\pi r^2[/tex]

Surface Area (A) of solid is

[tex]A = A_1 + A_2[/tex]

[tex]A = 2\pi rh + 4\pi r^2[/tex]

Substitute [tex]h = \frac{10}{\pi r^2} - \frac{4r }{3}[/tex]

[tex]A = 2\pi r(\frac{10}{\pi r^2} - \frac{4r }{3}) + 4\pi r^2[/tex]

Open bracket

[tex]A = \frac{2\pi r*10}{\pi r^2} - \frac{2\pi r*4r }{3} + 4\pi r^2[/tex]

[tex]A = \frac{2*10}{r} - \frac{2\pi r*4r }{3} + 4\pi r^2[/tex]

[tex]A = \frac{20}{r} - \frac{8\pi r^2 }{3} + 4\pi r^2[/tex]

[tex]A = \frac{20}{r} + \frac{-8\pi r^2 }{3} + 4\pi r^2[/tex]

Take LCM

[tex]A = \frac{20}{r} + \frac{-8\pi r^2 + 12\pi r^2}{3}[/tex]

[tex]A = \frac{20}{r} + \frac{4\pi r^2}{3}[/tex]

Differentiate w.r.t r

[tex]A' = -\frac{20}{r^2} + \frac{8\pi r}{3}[/tex]

Equate A' to 0

[tex]-\frac{20}{r^2} + \frac{8\pi r}{3} = 0[/tex]

Solve for r

[tex]\frac{8\pi r}{3} = \frac{20}{r^2}[/tex]

Cross Multiply

[tex]8\pi r * r^2 = 20 * 3[/tex]

[tex]8\pi r^3 = 60[/tex]

Divide both sides by [tex]8\pi[/tex]

[tex]r^3 = \frac{60}{8\pi}[/tex]

[tex]r^3 = \frac{15}{2\pi}[/tex]

Take [tex]\pi = 22/7[/tex]

[tex]r^3 = \frac{15}{2 * 22/7}[/tex]

[tex]r^3 = \frac{15}{44/7}[/tex]

[tex]r^3 = \frac{15*7}{44}[/tex]

[tex]r^3 = \frac{105}{44}[/tex]

Take cube roots of both sides

[tex]r = \sqrt[3]{\frac{105}{44}}[/tex]

[tex]r = \sqrt[3]{2.38636363636}[/tex]

[tex]r = 1.33632535155[/tex]

[tex]r = 1.34[/tex] (approximated)

Hence, the radius is 1.34cm

The radius of the cylinder that produces the minimum surface area is 1.34cm and this can be determined by using the formula area and volume of cylinder and hemisphere.

Given :

The volume of a cylinder is given by:

[tex]\rm V = \pi r^2 h[/tex]

The total volume of the two hemispheres is given by:

[tex]\rm V' = 2\times \dfrac{2}{3}\pi r^3[/tex]

[tex]\rm V' = \dfrac{4}{3}\pi r^3[/tex]

Now, the total volume of the solid is given by:

[tex]\rm V_T = \pi r^2 h+\dfrac{4}{3}\pi r^3[/tex]

Now, substitute the value of the total volume in the above expression and then solve for h.

[tex]\rm 10 = \pi r^2 h+\dfrac{4}{3}\pi r^3[/tex]

[tex]\rm h = \dfrac{10}{\pi r^2}-\dfrac{4r}{3}[/tex]

Now, the surface area of the curved surface is given by:

[tex]\rm A = 2\pi r h[/tex]

Now, the surface area of the two hemispheres is given by:

[tex]\rm A'=2\times (2\pi r^2)[/tex]

[tex]\rm A'=4\pi r^2[/tex]

Now, the total area is given by:

[tex]\rm A_T = 2\pi rh+4\pi r^2[/tex]

Now, substitute the value of 'h' in the above expression.

[tex]\rm A_T = 2\pi r\left(\dfrac{10}{\pi r^2}-\dfrac{4r}{3}\right)+4\pi r^2[/tex]

Simplify the above expression.

[tex]\rm A_T = \dfrac{20}{r} + \dfrac{4\pi r^2}{3}[/tex]

Now, differentiate the total area with respect to 'r'.

[tex]\rm \dfrac{dA_T}{dr} = -\dfrac{20}{r^2} + \dfrac{8\pi r}{3}[/tex]

Now, equate the above expression to zero.

[tex]\rm 0= -\dfrac{20}{r^2} + \dfrac{8\pi r}{3}[/tex]

Simplify the above expression in order to determine the value of 'r'.

[tex]8\pi r^3=60[/tex]

r = 1.34 cm

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