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Butoxors [25]
3 years ago
15

Use lagrange multiplier techniques to find the local extreme values of f(x, y) = x2 − y2 − 2 subject to the constraint x2 + y2 =

16
Mathematics
1 answer:
dexar [7]3 years ago
6 0
Given f(x,\ y)=x^2-y^2-2 subject to the constraint x^2+y^2=16

Let g(x,\ y)=x^2+y^2.

The gradient vectors of f and g are:

\nabla f(x,\ y)=\langle2x,-2y\rangle and \nabla g(x,\ y)=\langle2x,2y\rangle

By Lagrange's theorem, there is a number \lambda, such that

\langle2x,-2y\rangle=\lambda\langle2x,2y\rangle=\langle2\lambda x,2\lambda y\rangle

\lambda=\pm1

It can be seen that f(x,\ y)=x^2-y^2-2 has local extreme values at the given region.
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Nana76 [90]

Step-by-step explanation:

this is clearly not a linear sequence (the terms don't have the same difference).

so, it has to be a geometric sequence.

the common ratio is r.

s2 = s1 × r

16 = 64 × r

r = 16/64 = 1/4

control :

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scoundrel [369]

Answer:

All you have to do is divide both sides by 2.5 to find your answer.

2.5x / 2.5 = 35 / 2.5

x = 14

The answer is X = 14.

I hope this helps!

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