Mathematics: 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 =

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Use Lagrange multipliers to find the maximum and

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smortandsons

24.01.2022

Mathematics

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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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 =

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