Dhamnekar Win,odd Posted April 15 Share Posted April 15 Sufficient conditions for a critical point at (a, b, c) to be a local maximum or local minimum of a smooth function f(x, y, z).i-e [math]\nabla {f} (x,y,z)=0[/math]. How to define D in case of three variables namely x,y, z. Link to comment Share on other sites More sharing options...

Dhamnekar Win,odd Posted April 15 Author Share Posted April 15 (edited) More details about the question: The following theorem gives sufficient conditions for a critical point to be a local maximum or minimum of a smooth function (i.e. a function whose partial derivatives of all orders exist and are continuous). What are the changes, we must make in this theorem, in case f(x,y, z) and f(w, x , y, z)? How can we use the aforesaid rectified theorem to answer the following question? Find three positive numbers x, y, z whose sum is 10 such that [math] x^2 \cdot y^2 \cdot z[/math] is a maximum. My attempt : The critical point is x=4, y=4, z =2 How to compute D in this case? Edited April 15 by Dhamnekar Win,odd Link to comment Share on other sites More sharing options...

Genady Posted April 15 Share Posted April 15 3 hours ago, Dhamnekar Win,odd said: Find three positive numbers x, y, z whose sum is 10 such that x2⋅y2⋅z is a maximum. My attempt : The critical point is x=4, y=4, z =2 How to compute D in this case? x+y+z=10. You can express z by x and y. Then, you can substitute this expression for z in the given function, and it becomes a function of two variables. You can apply the theorem 2.6 as is to this function. 2 Link to comment Share on other sites More sharing options...

joigus Posted April 15 Share Posted April 15 (edited) 5 hours ago, Dhamnekar Win,odd said: How to compute D in this case? In the theorem's case, it's important to realise that, \[D=\frac{\partial^{2}f}{\partial x^{2}}\frac{\partial^{2}f}{\partial y^{2}}-\left(\frac{\partial^{2}f}{\partial x\partial y}\right)^{2}=\left|\begin{array}{cc} \frac{\partial^{2}f}{\partial x^{2}} & \frac{\partial^{2}f}{\partial x\partial y}\\ \frac{\partial^{2}f}{\partial y\partial x} & \frac{\partial^{2}f}{\partial y^{2}} \end{array}\right| \] In the case you propose, \[ \left|\begin{array}{ccc} \frac{\partial^{2}f}{\partial x^{2}} & \frac{\partial^{2}f}{\partial x\partial y} & \frac{\partial^{2}f}{\partial x\partial z}\\ \frac{\partial^{2}f}{\partial y\partial x} & \frac{\partial^{2}f}{\partial y^{2}} & \frac{\partial^{2}f}{\partial y\partial z}\\ \frac{\partial^{2}f}{\partial z\partial x} & \frac{\partial^{2}f}{\partial z\partial y} & \frac{\partial^{2}f}{\partial z^{2}} \end{array}\right| \] The condition that these 2nd-order determinants not being zero amount to being able to invert the relation that gives you the second-order derivatives and ascertaining the nature of your critical point. The determinant being zero amounts to second-order equations not being solvable, and therefore the method being inconclusive. There's an elegant method to solve this kind of problem which is the method of Lagrange multipliers. And there's the method that @Genady proposes too. Edited April 15 by joigus minor correction 1 Link to comment Share on other sites More sharing options...

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