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3 years ago

Help with math please? (View image then tell me answer and tell me how you got it from step-by-step)

Mathematics

1 answer:

77julia77 [94]3 years ago

5 0

Oh I remember these... just a lot of tedious work:/
but we are going to break the problem down by finding the volume of the half sphere and then the cylinder. After finding the measurements for both we just add them all together:)
first lets get the cylinder done
the formula for the can issss
$\pi r^{2} h$
so given that we know thatt
$\pi = 3.14$
r = 10
h = 16
so lets plug in our values
3.14 x 10² x 16
since they are all right by each other in the formula example we know that you multiply them. So lets solve that
3.14 x 10² x 16 = 5,024
So that means that the area of our cylinder is 5,024units³
Now before u go anywhere we still need to figure out the volume for the top:)
so the formula for our half sphere is
4/3 $\pi r$ ³
so lets find our value first
we know that 4/3 is just 4/3 soooo
4/3 = 4/3
$\pi = 3.14$
r = 10
so lets plug in our values
4/3 x 3.14 x 10³
now we just solve and 10³ = 1,000
4/3 x 3.14 x 1,000 = 4,186.66667units³
But since we are finding half of a sphere we are going to take that amount and divide it by two.
4,186.66667 ÷ 2 = 2,093.33334
So now we just add together our cylinder volume and half sphere
half sphere = 2,093.33334units³
cylinder = 5,024units³
2,093.33334 + 5,024 = 7,117.33334
soo the volume of our shape is 7,117.33334 or rounded out we could just say 7,117
answer = 7,117units³
hope this was helpful
and don't forget to
MARK ME BRIANLIEST! :D

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The plane x+y+2z=8 intersects the paraboloid z=x2+y2 in an ellipse. Find the points on this ellipse that are nearest to and fart

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The minimum distance of √((195-19√33)/8) occurs at ((-1+√33)/4; (-1+√33)/4; (17-√33)/4) and the maximum distance of √((195+19√33)/8) occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

Step-by-step explanation:

Here, the two constraints are

g (x, y, z) = x + y + 2z − 8

and

h (x, y, z) = x ² + y² − z.

Any critical point that we find during the Lagrange multiplier process will satisfy both of these constraints, so we actually don’t need to find an explicit equation for the ellipse that is their intersection.

Suppose that (x, y, z) is any point that satisfies both of the constraints (and hence is on the ellipse.)

Then the distance from (x, y, z) to the origin is given by

√((x − 0)² + (y − 0)² + (z − 0)² ).

This expression (and its partial derivatives) would be cumbersome to work with, so we will find the the extrema of the square of the distance. Thus, our objective function is

f(x, y, z) = x ² + y ² + z ²

and

∇f = (2x, 2y, 2z )

λ∇g = (λ, λ, 2λ)

µ∇h = (2µx, 2µy, −µ)

Thus the system we need to solve for (x, y, z) is

2x = λ + 2µx (1)

2y = λ + 2µy (2)

2z = 2λ − µ (3)

x + y + 2z = 8 (4)

x ² + y ² − z = 0 (5)

Subtracting (2) from (1) and factoring gives

2 (x − y) = 2µ (x − y)

so µ = 1 whenever x ≠ y. Substituting µ = 1 into (1) gives us λ = 0 and substituting µ = 1 and λ = 0 into (3) gives us 2z = −1 and thus z = − 1 /2 . Subtituting z = − 1 /2 into (4) and (5) gives us

x + y − 9 = 0

x ² + y ² + 1 /2 = 0

however, x ² + y ² + 1 /2 = 0 has no solution. Thus we must have x = y.

Since we now know x = y, (4) and (5) become

2x + 2z = 8

2x ² − z = 0

z = 4 − x

z = 2x²

Combining these together gives us 2x² = 4 − x , so

2x² + x − 4 = 0 which has solutions

x = (-1+√33)/4

and

x = -(1+√33)/4.

Further substitution yeilds the critical points

((-1+√33)/4; (-1+√33)/4; (17-√33)/4) and

(-(1+√33)/4; - (1+√33)/4; (17+√33)/4).

Substituting these into our objective function gives us

f((-1+√33)/4; (-1+√33)/4; (17-√33)/4) = (195-19√33)/8

f(-(1+√33)/4; - (1+√33)/4; (17+√33)/4) = (195+19√33)/8

Thus minimum distance of √((195-19√33)/8) occurs at ((-1+√33)/4; (-1+√33)/4; (17-√33)/4) and the maximum distance of √((195+19√33)/8) occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

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