Question

A cone shape cup is made from a circular piece of paper of radius 7 by cutting out a sector and joining the edge AC and BC. Find the maximum (volume) of such a cup

Answer 1

To find the maximum volume of the cone-shaped cup, we need to maximize the volume of the cone. The volume of a cone can be calculated using the formula V = (1/3)πr²h, where r is the radius of the circular base and h is the height of the cone.

Step-by-step explanation:

To find the maximum volume of the cone-shaped cup, we need to maximize the volume of the cone. The volume of a cone can be calculated using the formula V = (1/3)πr²h, where r is the radius of the circular base and h is the height of the cone.

In this case, the radius of the circular base is 7 units. To find the height of the cone, we need to determine the length of the slant height. The slant height is the hypotenuse of a right triangle formed by the radius and the height of the cone. Using the Pythagorean theorem, we can calculate the slant height as follows:

slant height² = radius² + height². Since the radius is 7 units and the length of the slant height is the same as the circumference of the base of the cone, the slant height is 2πr.

Substituting the values into the equation, we have:

(2πr)² = r² + h². Expanding and simplifying the equation gives:

4π²r² = r² + h². Rearranging the terms, we have:

h² = 4π²r² - r². Simplifying further, we get:

h = √(3π²r²).

Now, we can substitute the value of h in the volume formula and find the maximum volume:

V = (1/3)πr²√(3π²r²). Simplifying further, we get:

V = π(3πr⁴/9)^(1/2).

Answer 2

The answer is "138.10"

Step-by-step explanation:

Let r also be cone base radius and h = the cone height.

The radius for the original circle was on either side of the cone plied together = 7.

The hypotenuse is 7, as well as a right triangle, is formed. So:

`\to h^2+r^2 = 49\\\\ \to r^2 = 49-h^2`

Calculating the volume of the cone:

`V = ((1)/(3)) \pi r^2 h \\\\`

`= ((1)/(3)) \pi (49 -h^2) h\\\\= ((1)/(3)) \pi 49h -h^3\\\\`

`V'(h) = ((1)/(3)) \pi 49 - 3h^2\\\\`

Setting the derivative to zero and solving for h^2:

`49-3h^2 = 0 \\\\ h^2 = (49)/(3) \\\\ r^2 = 49-(49)/(3) \\`

`= ( 147 -49)/(3) \\\\= (98)/(3) \\\\`

Calculating the maximum volume:

`= (1)/(3) \pi (98)/(3) (7√(3))/(3) \\\\= (1)/(3) * 3.14 * 32.66 * 4.04 \\\\= (1)/(3) * 3.14 * 32.66 * 4.04 \\\\= 138.10`