What is the value of the interquartile range of the data below?
2 answers:
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Answer:
r = 3.667
h = 1.5
Step-by-step explanation:
Given:-
- The base radius of the right circular cone, R = 5.5
- The height of the right circular cone, H = 4.5
Solution:-
- We will first define two variables that identifies the volume of a cylinder as follows:
r: The radius of the cylinder
h: The height of cylinder
- Now we will write out the volume of the cylinder ( V ) as follows:
- We see that the volume of the cylinder ( V ) is a function of two variables ( don't know yet ) - ( r,h ). This is called a multi-variable function. However, some multi-variable functions can be reduced to explicit function of single variable.
- To convert a multi-variable function into a single variable function we need a relationship between the two variables ( r and h ).
- Inscribing, a cylinder in the right circular cone. We will denote 5 points.
Point A: The top vertex of the cone
Point B: The right end of the circular base ( projected triangle )
Point C: The center of both cylinder and base of cone.
Point D: The top-right intersection point of cone and cylinder
Point E: Denote the height of the cylinder on the axis of symmetry of both cylinder and cone.
- Now, we will look at a large triangle ( ABC ) and smaller triangle ( ADE ). We see that these two triangles are "similar". Therefore, we can apply the properties of similar triangles as follows:
- Now we can choose either variable variable to be expressed in terms of the other one. We will express the height of cylinder ( h ) in term of radius of cylinder ( r ) as follows:
- We will use the above derived relationship and substitute into the formula given above:
- Now our function of volume ( V ) is a single variable function. To maximize the volume of the cylinder we need to determine the critical points of the function as follows:
- We found the limiting value of the function. The cylinder volume maximizes when the radius ( r ) is two-thirds of the radius of the right circular cone.
- We can use the relationship between the ( r ) and ( h ) to determine the limiting value of height of cylinder as follows:
- The dimension of the inscribed cylinder with maximum volume are as follows:
Note: When we solved for the critical value of radius ( r ). We actually had two values: r = 0 , r = 2R/3. Where, r = 0 minimizes the volume and r = 2R/3 maximizes. Since the function is straightforward, we will not test for the nature of critical point ( second derivative test ).