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Svet_ta [14]
3 years ago
5

Use Lagrange multipliers to find the maximum and minimum values of f(x,y)=x+4y subject to the constraint x2+y2=9, if such values

exist. Round your answers to three decimal places. If there is no global maximum or global minimum, enter NA in the appropriate answer area. Maximum =
Mathematics
1 answer:
Vesnalui [34]3 years ago
4 0

The Lagrangian is

L(x,y,\lambda)=x+4y+\lambda(x^2+y^2-9)

with critical points where the partial derivatives vanish.

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

L_y=4+2\lambda y=0\implies y=-\dfrac2\lambda

L_\lambda=x^2+y^2-9=0

Substitute x,y into the last equation and solve for \lambda:

\left(-\dfrac1{2\lambda}\right)^2+\left(-\dfrac2\lambda\right)^2=9\implies\lambda=\pm\dfrac{\sqrt{17}}6

Then we get two critical points,

(x,y)=\left(-\dfrac3{\sqrt{17}},-\dfrac{12}{\sqrt{17}}\right)\text{ and }(x,y)=\left(\dfrac3{\sqrt{17}},\dfrac{12}{\sqrt{17}}\right)

We get an absolute maximum of 3\sqrt{17}\approx12.369 at the second point, and an absolute minimum of -3\sqrt{17}\approx-12.369 at the first point.

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