Optimisation Grade 12: Maximum Surface Area Cylinder
TLDRThe script outlines a mathematical problem to find the value of 'r' that maximizes the surface area of a cylinder. It begins by deriving the surface area formula, which includes the top and bottom circles and the lateral surface. With the given volume of 20 cubic centimeters, the height 'h' is expressed in terms of 'r'. Substituting 'h' into the surface area equation simplifies it to a function of 'r' alone. To find the maximum, the first derivative is taken and set to zero, leading to a cubic equation in 'r'. Solving this yields the optimal radius 'r' approximately equal to 1.47, which is the answer to the problem.
Takeaways
- 📚 The problem is to determine the value of 'r' for which the surface area of a cylinder is maximized.
- 🔍 The formula for the surface area of a cylinder includes the areas of the two circular bases and the lateral surface area.
- 🔢 The area of each base is given by \( \pi r^2 \), and since there are two bases, it's \( 2\pi r^2 \).
- 🌀 The lateral surface area is found by taking the circumference of the base, \( 2\pi r \), and multiplying it by the height 'h', resulting in \( 2\pi rh \).
- 🧩 The total surface area formula is thus \( 2\pi r^2 + 2\pi rh \).
- ⚖️ Given that the volume of the cylinder is 20 cubic centimeters, we use the volume formula \( \pi r^2h = 20 \) to express 'h' in terms of 'r'.
- 🔄 Substituting the expression for 'h' into the surface area formula simplifies it to \( 2\pi r^2 + \frac{40\pi r}{r^2} \).
- 📉 To find the maximum surface area, the first derivative of the simplified surface area formula with respect to 'r' is taken and set to zero.
- 📚 The first derivative simplifies to \( 4\pi r - \frac{40}{r^2} \), which is set to zero to solve for 'r'.
- 🔑 Solving the equation \( 4\pi r^3 = 40 \) gives the value of 'r' by dividing both sides by \( 4\pi \) and then taking the cube root.
- 🎯 The calculated value of 'r' is approximately 1.47, which is the radius that maximizes the surface area of the cylinder.
Q & A
What is the shape being discussed in the script?
-The shape being discussed is a cylinder.
What are the two areas that need to be painted on the outside of the cylinder?
-The two areas that need to be painted are the circular top and the circular bottom of the cylinder.
What is the formula for the area of a circle?
-The formula for the area of a circle is πr², where r is the radius.
How many circles are there in the total surface area calculation of a cylinder?
-There are two circles in the total surface area calculation, one at the top and one at the bottom of the cylinder.
What is the formula for the lateral surface area of a cylinder?
-The formula for the lateral surface area of a cylinder is 2πrh, where r is the radius and h is the height.
What is given as the volume of the cylinder in the script?
-The volume of the cylinder is given as 20 cubic centimeters.
What is the formula for the volume of a cylinder?
-The formula for the volume of a cylinder is πr²h, where r is the radius and h is the height.
How can the height (h) of the cylinder be expressed in terms of the radius (r) and the volume?
-The height (h) can be expressed as h = Volume / (πr²), which in this case is h = 20 / (πr²).
What is the purpose of taking the first derivative of the surface area formula?
-The purpose of taking the first derivative is to find the maximum or minimum values of the surface area by setting the derivative equal to zero.
What is the simplified formula for the surface area of the cylinder after substituting the volume formula?
-The simplified formula for the surface area becomes 2πr² + 40/r after substituting the volume formula.
How do you find the value of r that maximizes the surface area of the cylinder?
-To find the value of r that maximizes the surface area, you take the first derivative of the surface area formula, set it equal to zero, and solve for r.
What is the calculated value of r that maximizes the surface area of the cylinder?
-The calculated value of r that maximizes the surface area is approximately 1.47.
Outlines
📚 Calculating the Maximum Surface Area of a Cylinder
This paragraph discusses a mathematical problem involving finding the value of 'r' (radius) that maximizes the surface area of a cylinder. The speaker begins by explaining the formula for the surface area of a cylinder, which includes the areas of the two circular bases (2 * pi * r^2) and the lateral surface area (2 * pi * r * h). The challenge is to eliminate the variable 'h' (height), which is done by using the given volume of the cylinder (20 cubic centimeters) and the formula for the volume of a cylinder (pi * r^2 * h). By setting the volume equal to 20, 'h' is isolated and then substituted back into the surface area formula. The resulting equation is simplified to 40 * pi * r / r^2, which is then further manipulated to prepare for finding the maximum value by taking the first derivative and setting it to zero. The process involves algebraic manipulation to solve for 'r', which is found to be approximately 1.47.
Mindmap
Keywords
💡Surface Area
💡Cylinder
💡Pi (π)
💡Circumference
💡Volume
💡Base Area
💡Height (h)
💡First Derivative
💡Maximum
💡Optimization
💡Cube Root
Highlights
The question asks to determine the value of 'r' for which the surface area of a cylinder is maximized.
The formula for the surface area of a cylinder includes the areas of the top and bottom circles and the lateral surface area.
The area of a single circle is given by πr², and since there are two circles, it's 2πr².
The lateral surface area is calculated by taking the circumference of the circle (2πr) and multiplying it by the height (h).
The total surface area formula is 2πr² + 2πrh.
The volume of the cylinder is given as 20 cubic centimeters, and the volume formula is πr²h.
By equating the volume formula to 20, the height 'h' can be isolated and expressed in terms of 'r'.
Substituting the expression for 'h' into the surface area formula allows for a single-variable equation in terms of 'r'.
The simplified surface area formula becomes 2πr² + 40/r after substituting and simplifying.
To find maximums and minimums, the first derivative of the surface area formula with respect to 'r' is taken.
The first derivative simplifies to 4πr - 40/r².
Setting the first derivative equal to zero to find critical points for the surface area.
Solving the equation 4πr - 40/r² = 0 leads to a cubic equation in 'r'.
The solution to the cubic equation is found by isolating r³ and taking the cube root.
The calculated value of 'r' that maximizes the surface area is approximately 1.47.
The importance of verifying the solution against the original question's requirements is emphasized.
Transcripts
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