Answer:
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The points on the intersection of the ellipsoid with the plane that are respectively closest and furthest from the origin are
(2–√,−2–√,2−22–√)
(−2–√,2–√,2+22–√)
Using Lagrange multipliers we attempt to find the extrema of f(x,y,z)=x2+y2+z2 given that g(x,y,z)=x−y+z−2=0 and that h(x,y,z)=x2+y2−4=0.
Given,
∇f=⟨2x,2y,2z⟩
∇g=⟨1,−1,1⟩
∇h=⟨2x,2y,0⟩
Extrema satisfy the condition that ∇f=μ∇g+λ∇h for some λ,μ∈R.
This is to say,
2x=2λx+μ
2y=2λy−μ
2z=μ
If λ=1 then μ=0 and so z=0, the g constraint tells us that x=y+2, and the h constraint tells us that y2+(y+2)2=4, meaning that either y=0 or y=2. This provides us with two crucial points in addition to the g constraint:
(2,0,0)
(0,−2,0)
Now assume λ≠1, and so
x=μ2−2λ
y=−μ2−2λ=−x
Since x=−y, we have that x=±2–√, y=∓2–√. Using the g constraint, our two critical points are
(2–√,−2–√,2−22–√)
(−2–√,2–√,2+22–√)
And then it's east to determine which is the max and which is the min out of these four critical points.
To learn more about Lagrange multiplier
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Answer:
Required of volume of tank = 39.62 m3 or 10471.18 gallons
Step-by-step explanation:
Total volume of water collected = Annual Rainfall * Useful Area
Total volume of water collected = 1199.87 * 1.1667
Total volume of water collected = 1399.89 ft3
---1 ft3 = 7.48 gallons
Hence, total volume of water collected in gallons = 1399.89 * 7.48 = 10471.178 gallons
----1 ft3 = 1/35.315 m3 = 0.0283 m3
Therefore total volume of water collected in m3 = 1399.89 * 0.0283 = 39.6169 m3
Workings
Useful area = 61% of 1967
Useful area = 0.61 * 1967
Useful area = 1199.87 sq/ft.
Annual rainfall = 14 inch = 14/12 month
Annual rainfall = 1.1667 ft.
Answer:
87 adults
174 children
70 students
Step-by-step explanation:
Answer:
i don't have the answer but you can use the Pythagorean theorem to solve it when you know the hypotenuse and the angle. a2+b2=c2
Step-by-step explanation:
Some-one else posted this as a comment I just wanted it to be seen
