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+++ date = '2026-03-02T19:00:00-06:00' draft = false title = 'Function of Several Variables' layout = 'chapter' type = 'book' tags = 'calculus' chapterno = 15 +++
Max/Min Problems
Definition: [ \text{Suppose } (a,b) \text{ is a point in region} R \text{ on which } f \text{ is defined and there is an open disk centered at } (a,b) \text{.}\\ \text{If: } f(x,y) \leq f(a,b) \text{, then } f(a,b) \text{ is a local max.}\ \text{If: } f(x,y) \geq f(a,b) \text{, then } f(a,b) \text{ is a local min.} ]
Theorem: [ \text{If } f \text{ has a local max or min value at } (a,b) \text{ and the partial derivatives } f_x \text{ and } f_y \text{ exist at } (a,b) \text{, then } f_{x}(a,b) = f_{y}(a,b) = 0 \text{.} ]
If you are at ((a,b)) and every where you look, the value next to you is higher, you are at a local minimum position. If everywhere you look, the value is lower, you are at a local maximum position.
Critical points are identified by finding where the partial derivatives of each variable in the multi-variate function equals zero.
Definition: [ \text{ A critical point in } f \text{ is located as an interior point } (a,b) \text{ if either:}\\ f_{x}(a,b) = f_{y}(a,b) = 0\ f_{x} \text{ or } f{y} \text{ does not exist at } (a,b) ]
If you are able to find the critical points of (f) then you can use the Second Partial Derivative Test to determine the maximum and minimum values.
Theorem: [ \text{If the second partial derivatives of } f \text{ are continuous throughout an open disk centered at } (a,b) \text{. Let } D(x,y) = f_{xx}(x,y)f_{yy}(x,y) - (f_{xy}(x,y))^{2} \text{.}\\ \text{If } D(a,b) > 0 \text{ and } f_{xx}(a,b) < 0 \text{, there exists a local max at } (a,b) \text{.}\ \text{If } D(a,b) > 0 \text{ and } f_{xx}(a,b) < 0 \text{, there exists a local min at } (a,b) \text{.}\ \text{If } D(a,b) < 0 \text{, there exists a saddle point at }(a,b) \text{.}\ \text{If } D(a,b) = 0 \text{ the test is inconclusive.} ]
Lagrange Multipliers
Theorem: [ \text{Let } f \text{ be a differentiable function in } R^2 \text{ that contains curve } C \text{ given by } g(x,y) = 0. \text{ Assume } f \text{ has a local extrema on } C \text{ at point } P(a,b). \text{ Then } \nabla f(a,b) \text{ is orthogonal to the tangent line of } C \text{ at } P. \text{ Assuming } \nabla g(a,b) \neq 0, \text{ then there is a real number } \lambda \text{, or lagrange multiplier, such that } \nabla f(a,b) = \lambda \nabla g(a,b). ]
To find absolute extremas using lagrange multipliers:
- Find the gradient of (f)
- Find the gradient of (g)
- Set (\nabla f(x,y) = \lambda \nabla g(x,y))
- Solve the system for (x) and (y) of (\nabla f(x,y) = \lambda \nabla g(x,y)) and (g(x,y) = 0).
- Determine (x) and (y) for each case of (\lambda).
- Evaluate the points ((x,y)) you've found in (f(x,y)). The largest value is the absolute maximum. The smallest value is the absolute minimum.