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Calculus/Stationary points

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Expert: - 2/4/2010


Question
Hello Scotto, I am having trouble identifying the i) stationary points, and ii) the nature of the stationary points from the following question. Any help you could give would be very much appreciated.

Thank-You in advance.

f(x,y) = 3x^2 + 6xy + 2y^3 + 12x -24y

Answer
A stationary point will be where the slope is 0.
It can be seen that df/dx = 6x + 6y + 12 and df/dy = 6x + 6y² - 24.

Setting the the 1st equation equal to 0 gives 6x = -6y - 12.
Putting the 2nd equation equal 0 and putting that in gives 0 = -6y - 12 + 6y² - 24.
Combining like terms gives 6y² - 6y - 36 = 0.

Divide by 6. Note that y-3 cann factored out.  You can find the other term.

Set both factors equal to 0, so you have y-3=0 and the other term is 0.

These two equations can be solved for y.

Taking either of the derivatives as 0 and putting in these y values
will give an x value for each one.

Thus, you will have two points (x,y) that are the stationary points.

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