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Ch. 4 - Applications of the Derivative
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 4, Problem 4.5.71b

Cylinder in a cone A right circular cylinder is placed inside a cone of radius R and height H so that the base of the cylinder lies on the base of the cone.


b. Find the dimensions of the cylinder with maximum lateral surface area (area of the curved surface).

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1
Understand the problem: We need to find the dimensions of a cylinder with maximum lateral surface area that fits inside a cone with given radius R and height H. The lateral surface area of a cylinder is given by the formula: \( A = 2\pi rh \), where \( r \) is the radius and \( h \) is the height of the cylinder.
Set up the relationship between the cone and the cylinder: The cylinder is inscribed in the cone, so its dimensions are constrained by the cone's dimensions. The cone's equation in terms of its height and radius is \( \frac{r}{R} = \frac{H-h}{H} \), where \( r \) and \( h \) are the radius and height of the cylinder, respectively.
Express the height of the cylinder in terms of its radius: From the relationship \( \frac{r}{R} = \frac{H-h}{H} \), solve for \( h \) to get \( h = H - \frac{Hr}{R} \).
Substitute \( h \) in the lateral surface area formula: Replace \( h \) in \( A = 2\pi rh \) with the expression from the previous step to get \( A = 2\pi r(H - \frac{Hr}{R}) \). Simplify this to \( A = 2\pi rH - \frac{2\pi Hr^2}{R} \).
Find the maximum lateral surface area: To maximize \( A \), take the derivative of \( A \) with respect to \( r \), set it to zero, and solve for \( r \). This will give the radius of the cylinder that maximizes the lateral surface area. Use this \( r \) to find the corresponding \( h \) using the expression from step 3.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Lateral Surface Area of a Cylinder

The lateral surface area of a cylinder is the area of the curved surface that connects the top and bottom bases. It can be calculated using the formula A = 2πrh, where r is the radius of the base and h is the height of the cylinder. Understanding this concept is crucial for determining how to maximize the surface area in the context of the problem.
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Example 1: Minimizing Surface Area

Optimization in Calculus

Optimization involves finding the maximum or minimum values of a function within a given domain. In this problem, we need to apply techniques such as taking derivatives and setting them to zero to find the dimensions of the cylinder that maximize its lateral surface area. This concept is fundamental in calculus for solving real-world problems involving constraints.
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Intro to Applied Optimization: Maximizing Area

Geometric Relationships in 3D Shapes

Understanding the geometric relationships between the cylinder and the cone is essential for this problem. The dimensions of the cylinder are constrained by the dimensions of the cone, specifically its radius R and height H. This relationship can be expressed using similar triangles, which helps in deriving the equations needed to optimize the lateral surface area.
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Derivatives Applied To Velocity
Related Practice
Textbook Question

Pen problems


b. A rancher plans to make four identical and adjacent rectangular pens against a barn, each with an area of 100 m² (see figure). What are the dimensions of each pen that minimize the amount of fence that must be used? <IMAGE>

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Textbook Question

{Use of Tech} Basketball shot A basketball is shot with an initial velocity of v ft/s at an angle of 45° to the floor. The center of the basketball is 8 ft above the floor at a horizontal distance of 18 feet from the center of the basketball hoop when it is released. The height h (in feet) of the center of the basketball after it has traveled a horizontal distance of x feet is modeled by the function h(x) = 32x² / v² + x + 8 (see figure). <IMAGE>



b. During the flight of the basketball, show that the distance s from the center of the basketball to the front of the hoop is s = √ (x - 17.25)² + ( -(4x² / 81) + x - 2)² (Hint: The diameter of the basketball hoop is 18 inches.) 

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Textbook Question

Values of related functions Suppose f is differentiable on (-∞,∞) and assume it has a local extreme value at the point x = 2, where f(2) = 0. Let g(x) = xf(x) + 1 and let h(x) = xf(x) + x +1, for all values of x.


b. Does either g or h have a local extreme value at x = 2? Explain.

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Textbook Question

{Use of Tech} Demand functions and elasticity Economists use demand functions to describe how much of a commodity can be sold at varying prices. For example, the demand function D(p) = 500 - 10p says that at a price of p = 10, a quantity of D(10) = 400 units of the commodity can be sold. The elasticity E = dD/dp p/D of the demand gives the approximate percent change in the demand for every 1% change in the price. (See Section 3.6 or the Guided Project Elasticity in Economics for more on demand functions and elasticity.)


b. If the price is \$12 and increases by 4.5%, what is the approximate percent change in the demand? 

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{Use of Tech} Every second counts You must get from a point P on the straight shore of a lake to a stranded swimmer who is 50 from a point Q on the shore that is 50 m from you (see figure). Assuming that you can swim at a speed of 2 m/s and run at a speed of 4 m/s, the goal of this exercise is to determine the point along the shore, x meters from Q, where you should stop running and start swimming to reach the swimmer in the minimum time. <IMAGE>


b. Find the critical point of T on (0, 50).

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{Use of Tech} Let f(x) = ln((x+1)/(x-1)) and g(x) = ln ((x+1)/(x-1)).

b. Sketch graphs of f and g to show that these functions do not differ by a constant.

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