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Sufficient conditions for a critical point at (a, b, c)(i-e) [math]\nabla {f}(a,b,c)=0[/math]


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Posted (edited)

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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).

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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 by Dhamnekar Win,odd
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3 hours ago, Dhamnekar Win,odd said:

Find three positive numbers x, y, z whose sum is 10 such that x2y2z   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.

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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 by joigus
minor correction
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