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igor_vitrenko [27]
3 years ago
6

Pls help me 10 points

Mathematics
1 answer:
Reil [10]3 years ago
7 0

Answer:

1,1 and 3,-7

Step-by-step explanation:

1=x

1=y

they are both in positive

x=3

y=-7

here 3 is in positive and 7 is in negative

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Answer: 12.57 sq mm

Step-by-step explanation:

Area of a circle = pi r^2

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A cylinder is inscribed in a right circular cone of height 4.5 and radius (at the base) equal to 5.5 . What are the dimensions o
cluponka [151]

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:

                                V = \pi*r^2h

- 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:

                              \frac{AC}{AE} = \frac{BC}{DE}  \\\\\frac{H}{H-h} = \frac{R}{r}

- 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:

                             H- h = r\frac{H}{R} \\\\h = \frac{H}{R}*(R-r)

- We will use the above derived relationship and substitute into the formula given above:

                            V = \pi r^2 [ \frac{H}{R}*(R - r )]\\\\V = \frac{\pi H}{R}.r^2.(R-r)

- 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:

                            \frac{dV}{dr} =  \frac{\pi H}{R}*(2rR-2r^2 - r^2 )\\\\\frac{dV}{dr} =  \frac{\pi H}{R}*(2rR-3r^2 ) = 0\\\\(2rR-3r^2 ) = 0\\\\2R -3r = 0\\\\r = \frac{2}{3}*R

- 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:

                          h = \frac{H}{R} * ( R - \frac{2}{3}R)\\\\h = \frac{H}{3}

- The dimension of the inscribed cylinder with maximum volume are as follows:

                         r = \frac{2}{3}*5.5 = 3.667\\\\h = \frac{4.5}{3} = 1.5

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 ).

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