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Setler [38]
3 years ago
13

Use Lagrange multipliers to find the maximum and minimum values of

Mathematics
1 answer:
marusya05 [52]3 years ago
5 0

The Lagrangian is

L(x,y,\lambda)=x+8y+\lambda(x^2+y^2-4)

It has critical points where the first order derivatives vanish:

L_x=1+2\lambda x=0\implies\lambda=-\dfrac1{2x}

L_y=8+2\lambda y=0\implies\lambda=-\dfrac4y

L_\lambda=x^2+y^2-4=0

From the first two equations we get

-\dfrac1{2x}=-\dfrac4y\implies y=8x

Then

x^2+y^2=65x^2=4\implies x=\pm\dfrac2{\sqrt{65}}\implies y=\pm\dfrac{16}{\sqrt{65}}

At these critical points, we have

f\left(\dfrac2{\sqrt{65}},\dfrac{16}{\sqrt{65}}\right)=2\sqrt{65}\approx16.125 (maximum)

f\left(-\dfrac2{\sqrt{65}},-\dfrac{16}{\sqrt{65}}\right)=-2\sqrt{65}\approx-16.125 (minimum)

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Answer:

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Step-by-step explanation:

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Equation of the required line: y = mx + b

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