Question #332883

Use Lagrange Multipliers to find the maximum and minimum values of f(x,y)=xy

 subject to the constraint 4x^2+8y^2=16


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1
Expert's answer
2022-04-25T16:17:40-0400

f(x,y)=xy,  4x2+8y2=16.f(x,y)=xy, \ \ 4x^2+8y^2=16.

Let g(x,y)=4x2+8y2.g(x,y)=4x^2+8y^2.


We have fx=y,  fy=xf_x=y,\ \ f_y=x and gx=8x,  gy=16y.g_x=8x,\ \ g_y=16y.


{fx=λgxfy=λgy4x2+8y2=16{y=8λxx=16λy4x2+8y2=16\begin{cases} f_x=\lambda g_x \\ f_y=\lambda g_y \\ 4x^2+8y^2=16 \end{cases} \quad \begin{cases} y=8\lambda x \\ x=16\lambda y \\ 4x^2+8y^2=16 \end{cases}


x=16λy=16λ8λx=128λ2xx=16\lambda y=16\lambda \cdot8\lambda x=128\lambda ^2x

x0  128λ2=1,  λ=±182x\neq 0\ \Rightarrow \ 128\lambda^2=1,\ \ \lambda =\pm \tfrac{1}{8\sqrt{2}}


4x2+8y2=4x2+8(8λx)2=8x2=164x^2+8y^2=4x^2+8\cdot (8\lambda x)^2=8x^2=16

x=±2,x=\pm\sqrt{2}, y=±1y=\pm1


Maximum values: f(2,1)=f(2,1)=2f(\sqrt{2},1)=f(-\sqrt{2},-1)=\sqrt{2}

Minimum values:

f(2,1)=f(2,1)=2f(-\sqrt{2},1)=f(\sqrt{2},-1)=-\sqrt{2}


Answer: maximum value is 2\sqrt{2} , minimum value is 2-\sqrt{2} .



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