introduce concept of partial derivative and how to use in optimization of functions with 2 variables (a 3D surface function) and how to find the local max or min point of such 3D functions. use specific example in the end. also this is use for IB Math IA project. start your video with. Hi! This is IB45学长TUTOR, today we will introduce how we find the maximum and minimum of a multi-variable functions..."

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Hi! This is IB45学长TUTOR, today we will introduce how we find the maximum and minimum of multi-variable functions. Unlike one-dimensional functions that form curves on a two-dimensional plane, functions of two variables create surfaces in three-dimensional space. Our goal is to find the highest peaks or lowest valleys on these surfaces, which is essential for optimization problems in your IB Math IA projects. To find extrema on a three-dimensional surface, we need partial derivatives. A partial derivative with respect to x measures the slope in the x-direction, treating y as constant. Similarly, the partial derivative with respect to y measures the slope in the y-direction. Geometrically, these represent tangent line slopes parallel to the coordinate planes. To calculate them, simply treat other variables as constants and differentiate normally. To find extrema, we look for critical points where both partial derivatives equal zero simultaneously. This means the tangent plane at that point is horizontal with no slope in any direction. Critical points can be local maxima, local minima, or saddle points. In our example surface z equals x squared plus y squared minus one, the critical point occurs at the origin where both partial derivatives are zero, creating a horizontal tangent plane. To classify critical points, we use the second derivative test. We calculate three second partial derivatives: f sub xx, f sub yy, and f sub xy. Then we compute the discriminant D equals f sub xx times f sub yy minus f sub xy squared. If D is positive and f sub xx is positive, we have a local minimum. If D is positive and f sub xx is negative, we have a local maximum. If D is negative, we have a saddle point. These three types are shown in the visualization. Let's work through a complete example. For f of x y equals x squared plus y squared minus 6x plus 8y, we first find the partial derivatives: partial f partial x equals 2x minus 6, and partial f partial y equals 2y plus 8. Setting these equal to zero gives us x equals 3 and y equals negative 4. The second derivatives are f sub xx equals 2, f sub yy equals 2, and f sub xy equals 0. The discriminant D equals 4, which is positive, and f sub xx is positive, so we have a local minimum at point 3, negative 4 with function value negative 25. This method is essential for optimization problems in your IB Math IA projects.