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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 6, Problem 6.4.73b

Volumes without calculus Solve the following problems with and without calculus. A good picture helps.


b. A cube is inscribed in a right circular cone with a radius of 1 and a height of 3. What is the volume of the cube?

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First, visualize the problem by drawing a right circular cone with radius 1 and height 3, and then inscribe a cube inside it so that the cube touches the cone's sides and base.
Set up a coordinate system with the cone's vertex at the origin and the base at height 3. Express the radius of the cone's cross-section at any height \( y \) as a linear function: \( r(y) = 1 - \frac{y}{3} \).
Let the side length of the cube be \( s \). Position the cube so that its base lies on the cone's base (at \( y = 3 \)) and its top is at \( y = 3 - s \). The half-width of the cube at height \( y \) must be less than or equal to the radius of the cone at that height.
Use the relationship between the cube's half side length \( \frac{s}{2} \) and the cone's radius at the cube's top \( r(3 - s) \) to set up the equation \( \frac{s}{2} = r(3 - s) = 1 - \frac{3 - s}{3} \).
Solve this equation for \( s \) to find the side length of the cube, then compute the volume of the cube using \( V = s^3 \).

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Key Concepts

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

Inscribed Solids and Geometric Constraints

Understanding how a solid, like a cube, fits inside another shape, such as a cone, requires analyzing the geometric constraints. This involves relating dimensions of the inscribed solid to the dimensions of the outer shape, ensuring all vertices lie within or on the boundary of the cone.
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Geometric Sequences - Recursive Formula

Volume Formulas and Optimization

Calculating the volume of the cube involves using the formula V = s³, where s is the side length. To find the largest possible cube inside the cone, one must optimize s under the given constraints, often requiring setting up equations that relate s to the cone's radius and height.
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Using Calculus for Maximization Problems

Calculus techniques, such as derivatives, help find maximum or minimum values of functions. In this problem, calculus can be used to maximize the cube's volume by differentiating the volume function with respect to the cube's side length and solving for critical points.
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Related Practice
Textbook Question

6–8. Let R be the region bounded by the curves y = 2−√x,y=2, and x=4 in the first quadrant.

Suppose the shell method is used to determine the volume of the solid generated by revolving R about the line x=4.


b. What is the height of a cylindrical shell at a point x in [0, 4]?

Textbook Question

For the given regions R₁ and R₂, complete the following steps.


b. Find the area of region R₂ using geometry and the answer to part (a).


R₁is the region in the first quadrant bounded by the line x=1 and the curve y=6x(2−x^2)^2; R₂ is the region in the first quadrant bounded the curve y=6x(2−x^2)^2and the line y=6x.

Textbook Question

Displacement and distance from velocity Consider the velocity function shown below of an object moving along a line. Assume time is measured in seconds and distance is measured in meters. The areas of four regions bounded by the velocity curve and the t-axis are also given.

b. What is the displacement of the object over the interval [2, 6]? 

Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.


b. If a region is revolved about the y-axis, then the shell method must be used.

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Textbook Question

Emptying a water trough A water trough has a semicircular cross section with a radius of 0.25 m and a length of 3 m (see figure).

b. If the length is doubled, is the required work doubled? Explain.

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Textbook Question

55–58. Marginal cost Consider the following marginal cost functions.


b. Find the additional cost incurred in dollars when production is increased from 500 units to 550 units.


C′(x) = 300+10x−0.01x²