# Optimisation Grade 12: Maximum Volume Cylinder

TLDRThe video script discusses the process of finding the maximum volume of a cylinder given its surface area is 20. It explains the formula for volume and how to use the surface area to express height in terms of the radius. The script then guides through taking the first derivative of the volume with respect to the radius, setting it to zero to find the critical point, and solving for the radius that maximizes the volume. The final step involves simplifying the equation and concluding that the radius, r, should be approximately 1.03, but since radius cannot be negative, the correct value is 1.00.

###### Takeaways

- π The problem involves finding the maximum volume of a cylinder given a fixed surface area.
- π The volume formula for a cylinder is the area of the base times the height, which is \( \pi r^2 \times h \).
- π The surface area of a cylinder is given by \( 2\pi r^2 + 2\pi rh \), where \( 2\pi r^2 \) accounts for the top and bottom circles and \( 2\pi rh \) for the lateral surface.
- π’ The given surface area is 20, which allows us to derive a relationship between the radius \( r \) and height \( h \) of the cylinder.
- π§© By setting the surface area equation equal to 20, we can solve for \( h \) in terms of \( r \) to get \( h = \frac{10}{r} - 2r \).
- π The volume expression is then rewritten using the derived \( h \) to express volume solely in terms of \( r \), resulting in \( V = \pi r^2 \left(\frac{10}{r} - r\right) \).
- π Simplifying the volume formula leads to \( V = 10r - \pi r^3 \), which is a function of \( r \) only.
- π To find the maximum volume, we take the first derivative of the volume with respect to \( r \), resulting in \( \frac{dV}{dr} = 10 - 3\pi r^2 \).
- βοΈ Setting the derivative equal to zero to find critical points gives us \( 3\pi r^2 = 10 \).
- π Solving for \( r \) yields \( r = \sqrt{\frac{10}{3\pi}} \), which is approximately 1.03 when considering only the positive root since radius cannot be negative.
- π The final takeaway is that the radius \( r \) that maximizes the volume of the cylinder, given a surface area of 20, is approximately 1.03 units.

###### Q & A

### What is the main objective of the problem discussed in the video?

-The main objective of the problem is to find the value of 'r' that maximizes the volume of a cylinder.

### What is the formula for the volume of a cylinder in terms of its radius 'r' and height 'h'?

-The formula for the volume of a cylinder is the area of the base, which is Οr^2, multiplied by the height 'h'.

### Why is it necessary to use the surface area formula of a cylinder to solve this problem?

-The surface area formula is necessary because it provides a relationship between the radius 'r' and the height 'h', which allows us to express 'h' in terms of 'r' and thus simplify the volume formula to a single variable function.

### What is the given surface area of the cylinder in the problem?

-The given surface area of the cylinder is 20 square units.

### How is the height 'h' of the cylinder related to its radius 'r' after using the surface area formula?

-The height 'h' is related to the radius 'r' by the equation 2Οrh = 20 - 2Οr^2, which simplifies to h = 10/r after dividing by 2Οr.

### What is the simplified volume formula after substituting the expression for 'h' in terms of 'r'?

-The simplified volume formula is Οr^3(10/r - 1), which further simplifies to 10Οr^2 - Οr^3.

### What mathematical technique is used to find the maximum or minimum of a function with one variable?

-The technique used is to take the first derivative of the function and set it equal to zero to find the critical points, which may represent maximums or minimums.

### What is the first derivative of the simplified volume formula with respect to 'r'?

-The first derivative of the volume formula with respect to 'r' is 20Ο - 3Οr^2.

### How do you find the critical point for the volume formula?

-To find the critical point, set the first derivative equal to zero: 20Ο - 3Οr^2 = 0, and solve for 'r'.

### What is the value of 'r' that maximizes the volume of the cylinder according to the video?

-The value of 'r' that maximizes the volume is approximately 1.03, but since 'r' must be positive, the final answer is r = 1.03.

### Why can't the radius 'r' be negative in the context of this problem?

-The radius 'r' cannot be negative because a negative radius does not have a physical meaning in the context of a geometric shape like a cylinder.

###### Outlines

##### π Calculating Maximum Volume of a Cylinder

This paragraph discusses the process of finding the maximum volume of a cylinder given its surface area is 20. The speaker explains the basic formula for volume, which is the area of the base times the height, and notes the difference from a similar problem involving surface area. The formula for the volume of a cylinder is derived by substituting the height with an expression involving the radius 'r' and the surface area constraint. The resulting volume formula is simplified and then differentiated to find the critical points that may correspond to a maximum volume. The first derivative is set to zero, and the resulting equation is solved for 'r', leading to the conclusion that the radius 'r' should be approximately 1.03, but since radius cannot be negative, it is taken as 1.00.

###### Mindmap

###### Keywords

##### π‘Volume

##### π‘Surface Area

##### π‘Cylinder

##### π‘First Derivative

##### π‘Optimization

##### π‘Pi (Ο)

##### π‘Radius (r)

##### π‘Height (h)

##### π‘Derivative

##### π‘Square Root

##### π‘Maximization

###### Highlights

The question is about finding the value of r for maximum volume of a shape, contrasting with a previous problem on maximum surface area.

Volume formula is derived as area of base times height, applicable to shapes other than pyramids.

For a cylinder, the area of base is a circle with the formula ΟrΒ².

Cylinder's surface area is given as 20, leading to an equation involving r and h.

Surface area of a cylinder includes two circles and the circumference times height.

Derivation of h in terms of r using the surface area constraint 2Οrh = 20 - 2ΟrΒ².

Simplification of h to h = 10/r after canceling terms.

Substitution of h in the volume formula to express volume solely in terms of r.

Volume expression simplifies to 10r - ΟrΒ³, highlighting the dependency on r.

Application of calculus to find maximums by setting the first derivative equal to zero.

Derivative of the volume with respect to r results in 10 - 3ΟrΒ².

Solving the derivative equation for r to find critical points.

Critical value for r is calculated as the square root of 10/3Ο, considering only the positive root.

Final determination that r equals approximately 1.03, acknowledging the physical constraint of a positive radius.

The process emphasizes the importance of understanding the relationship between variables in optimization problems.

Demonstration of mathematical techniques to solve real-world geometry optimization problems.

The video serves as an educational tool for students learning calculus and its applications.

###### Transcripts

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