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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35 =
(3.14)(r²)(2.4)
35 =
(7.536)(r²)
35 = <span>2.512 r</span>²
35 ÷ 2.512 = 14
2.512 ÷ 2.512 = 1
14 ≈ r²
√14 ≈ √r²
3.74165 ≈ r
I'd say A would be the answer!
Answer:
-2
Step-by-step explanation:
the equation is y=2x-2 and the constant is the y-intercept
Answer:
JK = 13
Step-by-step explanation:
Note that corresponding sides are congruent.
Thus ST = KL and JK = RS
Given ST = 16RS and KL = 13JK + 39, thus
KL = 13JK + 39 ( KL = ST = 16RS ), substituting
16RS = 13RS + 39 ( JK = RS )
Subtract 13RS from both sides
3RS = 39 ( divide both sides by 3 )
RS = 13
However, RS and JK are corresponding sides and congruent, thus
JK = RS = 13
