Math Problem Statement

A solid cylinder of radius r cm and height h cm is to be carved out of a piece of solid wood in the form of a solid cone of base diameter 24 cm and height 32 cm. Calculate the value of r for which the volume of the cylinder carved out will be a maximum and find this maximum volume.

Solution

To solve this problem, we need to maximize the volume of a cylinder that can be carved out of a cone. Let's break down the solution step-by-step.

Step 1: Understand the Given Dimensions

  • The cone has a base diameter of 24 cm, so its radius R=242=12R = \frac{24}{2} = 12 cm.
  • The height of the cone H=32H = 32 cm.
  • The cylinder to be carved has a radius rr cm and a height hh cm.

Step 2: Set Up the Volume Formulas

  1. Volume of the Cone:

    The volume VconeV_{\text{cone}} of the cone is given by: Vcone=13πR2HV_{\text{cone}} = \frac{1}{3} \pi R^2 H Plugging in the values: Vcone=13π(12)2(32)=13π×144×32=1536π cm3V_{\text{cone}} = \frac{1}{3} \pi (12)^2 (32) = \frac{1}{3} \pi \times 144 \times 32 = 1536\pi \text{ cm}^3

  2. Volume of the Cylinder:

    The volume VcylinderV_{\text{cylinder}} of the cylinder is: Vcylinder=πr2hV_{\text{cylinder}} = \pi r^2 h

Step 3: Relate the Dimensions of the Cylinder and the Cone

The cylinder is inscribed in the cone, so its dimensions must fit within the cone. If the cylinder's radius is rr, and its height is hh, these dimensions are related to the cone's geometry.

Using similar triangles, the ratio of corresponding sides gives us: rR=HhH\frac{r}{R} = \frac{H - h}{H} r12=32h32\frac{r}{12} = \frac{32 - h}{32} Solving for hh: r=12(32h32)r = 12 \left( \frac{32 - h}{32} \right) r=1212h32r = 12 - \frac{12h}{32} r=123h8r = 12 - \frac{3h}{8}

Now, solve for hh in terms of rr: 3h8=12r\frac{3h}{8} = 12 - r h=8(12r)3h = \frac{8(12 - r)}{3} h=328r3h = 32 - \frac{8r}{3}

Step 4: Express the Volume of the Cylinder as a Function of rr

Substitute hh into the volume formula for the cylinder: Vcylinder=πr2(328r3)V_{\text{cylinder}} = \pi r^2 \left(32 - \frac{8r}{3}\right) Expand and simplify: Vcylinder=πr2(328r3)=π(32r28r33)V_{\text{cylinder}} = \pi r^2 \left(32 - \frac{8r}{3}\right) = \pi \left(32r^2 - \frac{8r^3}{3}\right) Vcylinder=π(32r283r3)V_{\text{cylinder}} = \pi \left(32r^2 - \frac{8}{3}r^3\right)

Step 5: Find the Value of rr for Maximum Volume

To find the maximum volume, we need to differentiate VcylinderV_{\text{cylinder}} with respect to rr and set the derivative equal to zero.

dVcylinderdr=π(64r8r2)=0\frac{dV_{\text{cylinder}}}{dr} = \pi \left(64r - 8r^2\right) = 0

Simplify: 64r8r2=064r - 8r^2 = 0 8r(8r)=08r(8 - r) = 0

So, r=0r = 0 or r=8r = 8.

Since r=0r = 0 would imply no cylinder, we choose r=8r = 8.

Step 6: Calculate the Maximum Volume

Substitute r=8r = 8 into the volume formula: h=328×83=32643=96643=323h = 32 - \frac{8 \times 8}{3} = 32 - \frac{64}{3} = \frac{96 - 64}{3} = \frac{32}{3}

Now, calculate VcylinderV_{\text{cylinder}}: Vcylinder=πr2h=π(82)(323)V_{\text{cylinder}} = \pi r^2 h = \pi (8^2) \left(\frac{32}{3}\right) Vcylinder=π(64)(323)=2048π3 cm3V_{\text{cylinder}} = \pi (64) \left(\frac{32}{3}\right) = \frac{2048\pi}{3} \text{ cm}^3

Conclusion

  • The value of rr for which the volume of the cylinder is maximized is r=8r = 8 cm.
  • The maximum volume of the cylinder is 2048π3\frac{2048\pi}{3} cm3^3.

Would you like any further details or have any questions?

Related Questions

  1. What is the surface area of the cylinder when r=8r = 8 cm and h=323h = \frac{32}{3} cm?
  2. How does the volume of the cylinder change if the cone's height is halved?
  3. What would be the maximum volume of a cylinder carved from a cone with a different base radius?
  4. How does the volume change if the radius rr is 6 cm instead of 8 cm?
  5. What is the relationship between the radius and height of the cylinder for it to have maximum volume?

Tip

When dealing with optimization problems, always start by defining the relationships between the variables and constraints to simplify the process of finding maxima or minima.

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Math Problem Analysis

Mathematical Concepts

Geometry
Optimization

Formulas

Volume of a Cone
Volume of a Cylinder

Theorems

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Suitable Grade Level

High School