Stationary Points of Quadratic Functions

Get partial derivatives in terms of x and y and z and etc

Make it equal to zero and find answer

The determinant of the matrix when calculating the answer(just the xyz,etc) part can be used to see if the stationary point is unique

This is the stationary point

Classifying the stationary point:

The equation can be made into matrix form using the quadratic portion of the equation. x^tAx like from before. A is a symmetric matrix. Use principal minor of A to find if point is positive definite. If positive definite then there can be a stationary point. If det of A is > 0 then minimum. If det A < 0 then maximum.

ab - c^2 > 0 minimum

ab-c^2 < 0 maximum

Can be applied to profit maximisation for firms with two products

Same as above just applied to this setting

Remember to multiply the prices by the quantities

Add the two revenues and take away the cost

Differentiate by both x and y and find the optimal point.

Classify the point.