anonymous
  • anonymous
Im not sure how to do this one: The temperature on the surface given by x^2 + y^2 + 4z^2 = 12 is given by T(x, y, z) = y^2 + 4z^2. What is the maximum temperature on this surface?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
You should use Lagrange multipliers to find the maximum temperature on the surface (which is an ellipsoid). The function is \[T(x,y,z)=y^2+4z^2.\] The constraint is \[g(x,y,z)=x^2+y^2+4z^2=12.\] The auxiliary function is \[\Lambda(x,y,z,\lambda)=T(x,y,z)+\lambda [g(x,y,z)-12]=\] \[=y^2+4z^2+\lambda[x^2+y^2+4z^2-12].\] Then you have to solve the following system of linear equations: \[\begin{array}{rcl} \frac{\partial\Lambda}{\partial x}=2x&=&0\\ \frac{\partial\Lambda}{\partial y}=2y+2\lambda y&=&0\\ \frac{\partial\Lambda}{\partial z}=8z+8\lambda z&=&0\\ \frac{\partial\Lambda}{\partial\lambda}=x^2+y^2+4z^2-12&=&0\\ \end{array}\] The solution is \[x=0,\quad \lambda=-1,\quad\textrm{ and you'll get: }y^2+4z^2=12.\] Since \[T(x,y,z)=y^2+4z^2\] therefore \[T(0,y,z)=12,\quad \textrm{ if } y=\pm\sqrt{12-4z^2},\quad -\sqrt{3}\leq z\leq \sqrt{3}.\] So the maximum temperature on the ellipsoid is 12 and the critical point are on the ellipse(!) \[y^2+4z^2=12.\]

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